Properties of alcohol Questions in English

(C) In a secondary benzylic alcohol,the $-OH$ group is attached to an $sp^3$ hybridized carbon atom,which is directly bonded to an aromatic ring and two other carbon atoms (or one carbon and one hydrogen atom,specifically a secondary carbon).
$1$. In option $A$,the $-OH$ group is attached to a primary carbon $(C_6H_5CH_2OH)$,making it a primary benzylic alcohol.
$2$. In option $B$,the $-OH$ group is attached to a tertiary carbon $(C_6H_5C(CH_3)_2OH)$,making it a tertiary benzylic alcohol.
$3$. In option $C$,the $-OH$ group is attached to a secondary carbon $(C_6H_5CH(OH)CH_3)$,which is bonded to the aromatic ring,making it a secondary benzylic alcohol.
$4$. In option $D$,the $-OH$ group is attached to a tertiary carbon $(C_6H_5C(OH)(CH_3)CH_2CH_3)$,making it a tertiary benzylic alcohol.
Therefore,the correct answer is $C$.

(B) benzylic alcohol is one where the $-OH$ group is attached to a carbon atom next to an aromatic ring.
$1$. In $C_6H_5CH_2OH$,the carbon attached to $-OH$ is bonded to one carbon (the ring) and two hydrogens,making it a primary $(1^{\circ})$ benzylic alcohol.
$2$. In $C_6H_5CH(OH)CH_3$,the carbon attached to $-OH$ is bonded to one carbon (the ring) and one methyl group,making it a secondary $(2^{\circ})$ benzylic alcohol.
$3$. In $C_6H_5C(CH_3)_2OH$,the carbon attached to $-OH$ is bonded to the ring and two methyl groups,making it a tertiary $(3^{\circ})$ benzylic alcohol.
$4$. In $C_6H_5C(CH_3)(C_2H_5)OH$,the carbon attached to $-OH$ is bonded to the ring,one methyl,and one ethyl group,making it a tertiary $(3^{\circ})$ benzylic alcohol.
Thus,the correct option is $B$.

(D) An allylic alcohol is one where the $-OH$ group is attached to a carbon atom adjacent to a carbon-carbon double bond $(C=C-C-OH)$.
$A$ tertiary alcohol is one where the carbon atom bearing the $-OH$ group is attached to three other carbon atoms.
In $2-$Methylbut$-3-$en$-2-$ol $(CH_2=CH-C(CH_3)_2-OH)$,the $-OH$ group is attached to a carbon that is adjacent to a double bond (making it allylic) and is bonded to three other carbon atoms (making it tertiary).

(B) The conversion of $2-$methylpropan$-1-$ol to $2-$methylpropan$-2-$ol is a rearrangement reaction.
In these reactions,the structure of the molecule is rearranged to form a structural isomer.
Specifically,this process involves the migration of the hydroxyl group or a hydride shift to form a more stable carbocation intermediate,which then leads to the formation of the tertiary alcohol,$2-$methylpropan$-2-$ol.

(C) Alcohols react with phosphorus pentachloride $(PCl_5)$ to produce alkyl chlorides,phosphorus oxychloride $(POCl_3)$,and hydrogen chloride $(HCl)$.
The balanced chemical equation is:
$R-OH + PCl_5 \longrightarrow R-Cl + POCl_3 + HCl$

(C) Step $1$: $Butan-2-ol$ reacts with $PCl_3$ to form $2-chlorobutane$ $(X)$.
$CH_3-CH_2-CH(OH)-CH_3 + PCl_3 \rightarrow CH_3-CH_2-CH(Cl)-CH_3 + H_3PO_3$
Step $2$: $2-chlorobutane$ $(X)$ undergoes dehydrohalogenation with alcoholic $KOH$ to form $But-2-ene$ $(Y)$ as the major product.
$CH_3-CH_2-CH(Cl)-CH_3 + alc. KOH \rightarrow CH_3-CH=CH-CH_3 (Y) + KCl + H_2O$
Step $3$: $But-2-ene$ $(Y)$ reacts with $H_2SO_4$ followed by hydrolysis $(H-OH/heat)$ to undergo hydration according to Markovnikov's rule,yielding $Butan-2-ol$ $(Z)$.
$CH_3-CH=CH-CH_3 + H_2SO_4 \rightarrow CH_3-CH_2-CH(OSO_3H)-CH_3$
$CH_3-CH_2-CH(OSO_3H)-CH_3 + H_2O \xrightarrow{\Delta} CH_3-CH_2-CH(OH)-CH_3 (Z) + H_2SO_4$

(A) The dehydration of $2-$Methylhexan$-3-$ol with concentrated $H_2SO_4$ proceeds via the formation of a carbocation intermediate.
Upon elimination of a proton,multiple alkenes can be formed.
According to the Saytzeff rule,the more substituted alkene is the major product.
$2-$Methylhexan$-3-$ol undergoes dehydration to form $2-$Methylhex$-2-$ene as the major product because it is a trisubstituted alkene,which is more stable than the disubstituted $2-$Methylhex$-3-$ene.

(A) Step $1$: Dehydration of $Propan-1-ol$ with $Al_2O_3$ at $623 \ K$ gives propene $(A)$ as the product.
$CH_3-CH_2-CH_2-OH \xrightarrow[623 \ K]{Al_2O_3} CH_3-CH=CH_2 (A)$
Step $2$: Addition of $conc. H_2SO_4$ to propene follows $Markovnikov's$ rule to form isopropyl hydrogen sulphate,which upon hydrolysis with water $(H_2O, \Delta)$ yields $Propan-2-ol$ $(B)$.
$CH_3-CH=CH_2$ $\xrightarrow{conc. H_2SO_4} CH_3-CH(OSO_3H)-CH_3$ $\xrightarrow{H_2O, \Delta} CH_3-CH(OH)-CH_3 (B)$
Thus,the final product $B$ is $Propan-2-ol$.

(B) Step $1$: Dehydration of $CH_3-CH_2-CH_2-OH$ (Propan$-1-$ol) with $Al_2 O_3$ at $623 \ K$ gives $A$,which is Propene $(CH_3-CH=CH_2)$.
Step $2$: Propene reacts with conc. $H_2 SO_4$ followed by hydrolysis $(H_2 O)$ to undergo hydration according to Markovnikov's rule.
Step $3$: The reaction follows the mechanism: $CH_3-CH=CH_2 + H_2 SO_4$ $\rightarrow CH_3-CH(OSO_3 H)-CH_3$ $\xrightarrow{H_2 O} CH_3-CH(OH)-CH_3$ (Propan$-2-$ol).
Thus,the product $B$ is Propan$-2-$ol.

(B) Dehydration of alcohols involves the formation of carbocation intermediates. The ease of dehydration depends on the stability of the carbocation formed.
Since the stability order of carbocations is $3^{\circ} > 2^{\circ} > 1^{\circ}$,the order of ease of dehydration of alcohols is $3^{\circ} > 2^{\circ} > 1^{\circ}$.

(B) The acid-catalyzed dehydration of butan-$1$-ol $(CH_3-CH_2-CH_2-CH_2OH)$ involves the formation of a primary carbocation,which undergoes a $1,2$-hydride shift to form a more stable secondary carbocation $(CH_3-CH^+-CH_2-CH_3)$.
Elimination of a proton from this secondary carbocation yields but-$2$-ene $(CH_3-CH=CH-CH_3)$ as the major product,following Zaitsev's rule,as it is more substituted and thus more stable than but-$1$-ene $(CH_3-CH_2-CH=CH_2)$.

(D) The reaction of alcohols with $Cu$ at $573 \ K$ is a dehydrogenation reaction,not dehydration.
$1$. Primary alcohols $(RCH_2OH)$ form aldehydes.
$2$. Secondary alcohols $(R_2CHOH)$ form ketones.
$3$. Tertiary alcohols $(R_3COH)$ undergo dehydration to form alkenes because they cannot undergo dehydrogenation due to the absence of $\alpha$-hydrogen.
Since the question asks for the reaction with $Cu$ at $573 \ K$,tertiary alcohols specifically undergo dehydration under these conditions. Therefore,the correct option is $D$.

(A) The reaction of a Grignard reagent $(R-MgX)$ with methanal $(HCHO)$ followed by hydrolysis yields a primary alcohol with one more carbon atom than the Grignard reagent.
To obtain $2-$methylpropan$-1-$ol $(CH_3-CH(CH_3)-CH_2-OH)$,the Grignard reagent must be isobutylmagnesium halide $(CH_3-CH(CH_3)-CH_2-MgX)$.
The reaction is:
$CH_3-CH(CH_3)-CH_2-MgX + HCHO \rightarrow CH_3-CH(CH_3)-CH_2-CH_2-OMgX$
$CH_3-CH(CH_3)-CH_2-CH_2-OMgX + H_2O \rightarrow CH_3-CH(CH_3)-CH_2-CH_2-OH + Mg(OH)X$
Wait,the target product is $2-$methylpropan$-1-$ol $(C_4H_{10}O)$. The reaction of $R-MgX + HCHO$ adds one carbon. Thus,$R$ must be an isopropyl group $(CH_3-CH(CH_3)-)$.
$CH_3-CH(CH_3)-MgX + HCHO$ $\rightarrow CH_3-CH(CH_3)-CH_2-OMgX$ $\xrightarrow{H_2O} CH_3-CH(CH_3)-CH_2-OH$.
Therefore,the correct Grignard reagent is isopropylmagnesium halide,which is option $A$.

(A) The boiling point of organic compounds depends on the intermolecular forces present.
$1$. Pentan-$1$-ol is an alcohol,which exhibits strong intermolecular hydrogen bonding.
$2$. Pentanal is an aldehyde,which has dipole-dipole interactions.
$3$. Ethoxy ethane is an ether,which has weaker dipole-dipole interactions.
$4$. $n$-Butane is an alkane,which only has weak London dispersion forces.
Since hydrogen bonding is the strongest intermolecular force among these,Pentan-$1$-ol has the highest boiling point.

(D) The boiling point of alcohols depends on the extent of hydrogen bonding and the surface area of the molecule.
$1$. Among the given options,$butan-1-ol$ is a primary alcohol with a straight-chain structure,which provides a larger surface area for van der Waals forces compared to branched isomers.
$2$. $Propan-2-ol$ has $3$ carbons,$butan-2-ol$ is a secondary alcohol,and $2-Methylpropan-2-ol$ is a tertiary alcohol.
$3$. As branching increases,the surface area decreases,leading to weaker intermolecular forces and lower boiling points.
$4$. Therefore,$butan-1-ol$ has the highest boiling point among the given options due to its linear structure and higher molecular mass compared to $propan-2-ol$.

(B) The reaction between a Grignard reagent $(CH_3CH_2MgBr)$ and a ketone $(CH_3COCH_3)$ proceeds via nucleophilic addition to the carbonyl group.
$1$. The ethyl group $(CH_3CH_2^-)$ acts as a nucleophile and attacks the electrophilic carbonyl carbon of propanone.
$2$. This forms an intermediate magnesium alkoxide: $(CH_3)_2C(OMgBr)(CH_2CH_3)$.
$3$. Subsequent acid hydrolysis of this intermediate yields the tertiary alcohol: $(CH_3)_2C(OH)(CH_2CH_3)$.
$4$. The $IUPAC$ name of the product $(CH_3)_2C(OH)CH_2CH_3$ is $2-$methylbutan$-2-$ol.
Therefore,the correct option is $B$.

(B) The products of the reactions are:
$A$. $CH_3-CH_2-CH_2OH$ (propan$-1-$ol,boiling point $\approx 97 \ ^\circ C$)
$B$. $CH_3-CH(OH)-CH_3$ (propan$-2-$ol,boiling point $\approx 82 \ ^\circ C$)
$C$. $CH_3-CH_2-CH_2OH$ (propan$-1-$ol,boiling point $\approx 97 \ ^\circ C$)
$D$. $CH_3-CH_2-CH_2OH$ (propan$-1-$ol,boiling point $\approx 97 \ ^\circ C$)
Among the isomers,branched-chain alcohols have a lower boiling point than straight-chain alcohols due to a smaller surface area,which leads to weaker van der Waals forces.
Therefore,propan$-2-$ol $(B)$ has the lowest boiling point.

(A) Primary alcohols $(R-CH_2OH)$ are oxidized to aldehydes $(R-CHO)$ using mild oxidizing agents.
$PCC$ (Pyridinium chlorochromate) in $CH_2Cl_2$ is a selective oxidizing agent that stops the oxidation at the aldehyde stage.
Strong oxidizing agents like $KMnO_4$ (in acidic or basic medium) or $Na_2Cr_2O_7 + H_2SO_4$ (Jones reagent) oxidize primary alcohols directly to carboxylic acids $(R-COOH)$.

(A) In the oxidation reaction of alcohols,the $O-H$ bond is cleaved,and a $C-H$ bond is broken to form a carbonyl group $(C=O)$. The $C-O$ bond remains intact.
In dehydration,$PBr_3$ reaction,and reaction with $HCl$ (Lucas test),the $C-O$ bond is cleaved to form alkenes or alkyl halides.
Therefore,the correct answer is $A$.

(D) The solubility of alcohols in water depends on the ability to form hydrogen bonds with water molecules.
As the number of hydroxyl $(-OH)$ groups increases,the number of hydrogen bonds formed with water increases,leading to higher solubility.
$A$: Secondary butyl alcohol $(C_4H_9OH)$ has $1$ $-OH$ group.
$B$: Tertiary butyl alcohol $(C_4H_9OH)$ has $1$ $-OH$ group.
$C$: Ethylene glycol $(HO-CH_2-CH_2-OH)$ has $2$ $-OH$ groups.
$D$: Glycerol $(HO-CH_2-CH(OH)-CH_2-OH)$ has $3$ $-OH$ groups.
Since glycerol has the maximum number of $-OH$ groups,it forms the most extensive hydrogen bonding with water,making it the most soluble among the given options.
Therefore,the correct option is $D$.

(D) The acidity of alcohols depends on the stability of the alkoxide ion formed after the loss of a proton $(H^+)$.
Alkyl groups are electron-donating groups ($+I$ effect).
As the number of alkyl groups attached to the carbon atom bearing the $-OH$ group increases,the electron density on the oxygen atom increases due to the $+I$ effect,which destabilizes the alkoxide ion.
Therefore,the order of acidity is: $primary > secondary > tertiary$.
Among the given options:
$(A)$ $(CH_3)_3C-OH$ is a tertiary alcohol.
$(B)$ $CH_3-CH_2-CH_2-OH$ is a primary alcohol.
$(C)$ $CH_3-CH(OH)-CH_3$ is a secondary alcohol.
$(D)$ $CH_3-OH$ is methanol (the simplest alcohol with no alkyl group attached to the carbon bearing $-OH$).
Since methanol has no electron-donating alkyl groups to destabilize the methoxide ion,it is the most acidic among the given options.

(C) The reactivity of alcohols with Lucas's reagent $(conc. \ HCl + anhydrous \ ZnCl_2)$ follows the order: $3^\circ > 2^\circ > 1^\circ$.
This is because the reaction proceeds via an $S_N1$ mechanism involving the formation of a carbocation intermediate.
$3^\circ$ alcohols form the most stable carbocation and thus react the fastest,typically at room temperature.
$(CH_3)_3COH$ is a tertiary $(3^\circ)$ alcohol,while the others are secondary $(2^\circ)$ or primary $(1^\circ)$ alcohols.
Therefore,$(CH_3)_3COH$ gives the fastest reaction.

(D) The reaction with sodium hypoiodite $(NaOI)$ is the iodoform test.
Compounds containing the $CH_3CH(OH)-$ group or the $CH_3CO-$ group give a positive iodoform test,resulting in a yellow precipitate of iodoform $(CHI_3)$.
Among the given options,$sec-$butyl alcohol $(CH_3CH(OH)CH_2CH_3)$ contains the $CH_3CH(OH)-$ structural unit.
Therefore,$sec-$butyl alcohol reacts with $NaOI$ to form a yellow precipitate.

(C) The oxidation of secondary alcohols like cyclohexanol to ketones like cyclohexanone can be achieved using various oxidizing agents.
Anhydrous $CrO_3$ (Jones reagent or similar chromium-based oxidants) is a standard reagent used for the oxidation of secondary alcohols to ketones.
Therefore,the correct reagent is anhydrous $CrO_3$.

(B) Denatured alcohol is ethyl alcohol that has been made unfit for drinking by adding poisonous substances such as methanol,naphtha,pyridine,or rubber. This process is done to prevent the misuse of industrial alcohol for consumption.

(D) $PCC$ stands for pyridinium chlorochromate.
It is a complex of chromium trioxide with pyridine and $HCl$,represented as $[C_5H_5NH]^{+}[CrO_3Cl]^{-}$.
It is a reagent in organic synthesis used primarily for the oxidation of primary alcohols to aldehydes and secondary alcohols to ketones.

(D) The reaction of ethanol with excess concentrated $H_2SO_4$ at $443 \ K$ is a dehydration reaction.
In this process,a molecule of water is removed from ethanol to form ethene.
The chemical equation is:
$CH_3CH_2OH \xrightarrow[443 \ K]{\text{Conc. } H_2SO_4} CH_2=CH_2 + H_2O$

(C) The conversion of propan$-2-$ol to propan$-1-$ol proceeds as follows:
$1$. Propan$-2-$ol reacts with $PCl_{5}$ to form $2-$chloropropane.
$2$. $2-$chloropropane undergoes dehydrohalogenation with alcoholic $KOH$ to form propene.
$3$. Propene undergoes hydroboration-oxidation (reaction with $B_{2}H_{6}$ followed by $H_{2}O_{2}/NaOH$) to form propan$-1-$ol via anti-Markovnikov addition of water.

(B)
$1$. Ethanol is a very weak acid,much weaker than water. It does not react with bases like $NaHCO_{3}$ or $Na_{2}CO_{3}$ and does not change the color of litmus paper.
$2$. However,it reacts with active metals like sodium $(Na)$ to evolve hydrogen gas $(H_{2})$,which confirms its acidic hydrogen atom.
$3$. The reaction is: $2C_{2}H_{5}OH + 2Na \rightarrow 2C_{2}H_{5}ONa + H_{2} \uparrow$

(A) When vapours of tertiary alcohols are passed over heated copper at $573 \ K$,they undergo dehydration to form alkenes.
$(CH_3)_3COH \xrightarrow[573 \ K]{Cu} CH_2=C(CH_3)_2 + H_2O$
Thus,the product formed is $2-$methylpropene.

(C) Lucas reagent is a mixture of concentrated $HCl$ and anhydrous $ZnCl_2$.
Tertiary $(3^{\circ})$ alcohols react immediately with Lucas reagent at room temperature to form alkyl chlorides,which appear as turbidity.
Secondary $(2^{\circ})$ alcohols react within $5-10$ minutes.
Primary $(1^{\circ})$ alcohols do not react at room temperature.
Among the given options,$2-$methyl propan$-2-$ol is a tertiary alcohol,so it gives turbidity immediately.

(D) The Lucas test is a chemical test used to distinguish between primary,secondary,and tertiary alcohols. It involves the reaction of an alcohol with Lucas reagent ($conc. \ HCl$ and $ZnCl_2$). Tertiary alcohols react immediately to form a cloudy precipitate,secondary alcohols react within $5-10 \ minutes$,and primary alcohols do not react at room temperature.

(A) The dehydration of ethanol to form ethoxyethane (diethyl ether) is an intermolecular dehydration reaction.
When an excess of ethanol is heated with concentrated $H_2SO_4$ at $413\ K$ $(140^{\circ}C)$,two molecules of ethanol undergo condensation to form ethoxyethane and water.
The reaction is: $2CH_3CH_2OH \xrightarrow{conc. H_2SO_4, 413\ K} CH_3CH_2-O-CH_2CH_3 + H_2O$.
Heating at $443\ K$ leads to intramolecular dehydration,forming ethene $(CH_2=CH_2)$.

(B) When vapors of a primary alcohol are passed over reduced copper catalyst at $573 \ K$,dehydrogenation occurs to form an aldehyde.
The reaction is represented as:
$R-CH_2OH \xrightarrow{Cu, 573 \ K} R-CHO + H_2$
Thus,the correct product is an aldehyde.

(B) When vapors of secondary alcohols are passed over heated copper $(Cu)$ catalyst at $573 \ K$,they undergo dehydrogenation to form ketones.
The general reaction is:
$R_2CH-OH \xrightarrow{Cu, 573 \ K} R_2C=O + H_2$

(B) The reaction of monohydric alcohol $(R-OH)$ with ethyl magnesium iodide $(C_2H_5MgI)$ is: $R-OH + C_2H_5MgI \rightarrow R-OMgI + C_2H_6 \uparrow$.
At $STP$,$1 \ mol$ of ethane occupies $22400 \ cm^3$.
Given $2240 \ cm^3$ of ethane is liberated,which corresponds to $2240 / 22400 = 0.1 \ mol$ of ethane.
Since $1 \ mol$ of alcohol produces $1 \ mol$ of ethane,$0.1 \ mol$ of alcohol has a mass of $8.8 \ g$.
Thus,the molar mass of the alcohol is $8.8 / 0.1 = 88 \ g/mol$.
The general formula for a saturated monohydric alcohol is $C_nH_{2n+1}OH$,so $12n + (2n+1) + 16 + 1 = 88$,which gives $14n = 70$,so $n = 5$.
The alcohol is a pentanol isomer $(C_5H_{12}O)$.
Since the alcohol forms an aldehyde upon oxidation with $PCC$ (which gives a positive Tollen's test),it must be a primary $(1^{\circ})$ alcohol.
Among the options,$2, 2$-dimethylpropan-$1$-ol has the formula $(CH_3)_3CCH_2OH$,which is a primary alcohol with molar mass $88 \ g/mol$.

(A) $PCC$ (pyridinium chlorochromate) is a commonly used reagent for the oxidation of primary alcohols to aldehydes.
It is favored due to its selectivity,mild reaction conditions,compatibility with other functional groups,and its ability to stop the oxidation at the aldehyde stage without further oxidation to carboxylic acids.

(B) $1$. Reaction of $CH_3CHO$ with $CH_3MgBr$ followed by hydrolysis gives $A$,which is $CH_3-CH(OH)-CH_3$ (Propan$-2-$ol).
$2$. Dehydration of $A$ with $Conc. H_2SO_4$ at high temperature gives $B$,which is $CH_3-CH=CH_2$ (Prop$-1-$ene).
$3$. Hydroboration-oxidation of $B$ $(CH_3-CH=CH_2)$ with $(i) B_2H_6$ and $(ii) H_2O_2/OH^-$ follows anti-Markovnikov addition of water to give $C$,which is $CH_3-CH_2-CH_2OH$ (Propan$-1-$ol).
$4$. Comparing $A$ $(CH_3-CH(OH)-CH_3)$ and $C$ $(CH_3-CH_2-CH_2OH)$,the position of the $-OH$ group is different (carbon-$2$ vs carbon-$1$). Therefore,they are position isomers.

(C) Compound $(Y)$ gives the iodoform test,which means it contains an acetyl group $(CH_{3}-C=O)$.
$PCC$ (Pyridinium chlorochromate) is an oxidizing agent that oxidizes primary alcohols to aldehydes and secondary alcohols to ketones.
Since compound $Y$ is formed from $X$ using $PCC$ and $Y$ gives the iodoform test,$Y$ must be acetaldehyde $(CH_{3}CHO)$,as it is the only aldehyde that gives the iodoform test.
Therefore,the starting material $X$ must be ethanol $(CH_{3}CH_{2}OH)$.
The reactions are as follows:
$CH_{3}CH_{2}OH$ $\xrightarrow{PCC} CH_{3}CHO$ $\xrightarrow{I_{2}/OH^{-}} CHI_{3} + HCOONa$

(C) Iodoform test is given by compounds containing the $-COCH_{3}$ group or the $-CH(OH)CH_{3}$ group.
Propan$-2-$ol contains the $-CH(OH)CH_{3}$ group.
Butan$-2-$one contains the $-COCH_{3}$ group.
Acetophenone contains the $-COCH_{3}$ group attached to a benzene ring.
Propan$-1-$ol is a primary alcohol $(CH_{3}CH_{2}CH_{2}OH)$ and does not contain the $-CH(OH)CH_{3}$ group. Ethanol is the only primary alcohol that gives the iodoform reaction.

(D) The oil of wintergreen test is an esterification reaction used to identify salicylic acid.
Salicylic acid reacts with methanol $(CH_{3}OH)$ in the presence of a few drops of concentrated $H_{2}SO_{4}$ (acid catalyst) upon heating to form methyl salicylate.
Methyl salicylate is an ester that has the characteristic smell of oil of wintergreen.
The reaction is: $C_{6}H_{4}(OH)COOH + CH_{3}OH \xrightarrow[Conc. H_{2}SO_{4}]{\Delta} C_{6}H_{4}(OH)COOCH_{3} + H_{2}O$.

(B) The iodoform reaction is given by compounds containing the $CH_{3}CO-$ group or the $CH_{3}CH(OH)-$ group.
$A) \ CH_{3}CHO$ contains the $CH_{3}CO-$ group,so it gives the iodoform reaction.
$B) \ CH_{3}CH_{2}CH_{2}OH$ (propan$-1-$ol) is a primary alcohol that does not contain the $CH_{3}CH(OH)-$ group. Therefore,it does not give the iodoform reaction.
$C) \ CH_{3}CH(OH)CH_{2}COOH$ contains the $CH_{3}CH(OH)-$ group,so it gives the iodoform reaction.
$D) \ CH_{3}CH_{2}OH$ (ethanol) is the only primary alcohol that contains the $CH_{3}CH(OH)-$ group,so it gives the iodoform reaction.
Thus,the correct answer is $B$.

(B) Acidified $KMnO_4$ is a strong oxidizing agent. It oxidizes primary alcohols directly to carboxylic acids rather than stopping at the aldehyde stage. Therefore,it cannot be used to prepare aldehydes from primary alcohols.

(C) The oxidation of a secondary alcohol using acidified $K_{2}Cr_{2}O_{7}$ yields a ketone.
Ketones react with phenyl hydrazine to form phenylhydrazones but do not give the silver mirror test (Tollens' test),which is specific to aldehydes.
Among the given options,$(CH_{3})_{2}CHOH$ is a secondary alcohol (propan$-2-$ol),which on oxidation gives acetone $(CH_{3}COCH_{3})$,a ketone.
Therefore,the correct structure of $X$ is $(CH_{3})_{2}CHOH$.

(B) The oxidation of a primary alcohol $(R-CH_{2}OH)$ to a carboxylic acid $(R-COOH)$ involves the replacement of two hydrogen atoms with one oxygen atom.
The mass difference is calculated as:
Mass of $-COOH$ group = $12 + 16 + 16 + 1 = 45 \ u$.
Mass of $-CH_{2}OH$ group = $12 + 2(1) + 16 + 1 = 31 \ u$.
Difference = $45 - 31 = 14 \ u$.
Thus,the compound is a primary alcohol.

(C) The best method for the conversion of an alcohol into an alkyl chloride is by treating the alcohol with $SOCl_{2}$ in the presence of pyridine.
$ROH + SOCl_{2} \longrightarrow RCl + SO_{2} \uparrow + HCl \uparrow$
In this reaction,the by-products $SO_{2}$ and $HCl$ are gases,which escape from the reaction mixture,leaving behind pure alkyl chloride. This is known as the Darzens process.

(D) The reaction sequence is as follows:
$1$. $\text{Ethanol} (CH_3 CH_2 OH)$ reacts with $PCl_5$ to form $X$,which is chloroethane $(CH_3 CH_2 Cl)$.
$2$. Chloroethane $(CH_3 CH_2 Cl)$ reacts with alcoholic $KOH$ (dehydrohalogenation) to form $Y$,which is ethene $(CH_2 = CH_2)$.
$3$. Ethene $(CH_2 = CH_2)$ reacts with $H_2SO_4$ at room temperature followed by hydrolysis $(H_2O, \Delta)$ to form $Z$,which is ethanol $(CH_3 CH_2 OH)$.
Therefore,the product $Z$ is $CH_3 CH_2 OH$.

(A) At $S.T.P.$,$22400 \ cm^{3}$ of methane corresponds to $1 \ mol$.
Therefore,$112 \ cm^{3}$ of methane corresponds to $\frac{112}{22400} = 0.005 \ mol$.
Since $1 \ mol$ of monohydric alcohol reacts with $1 \ mol$ of methylmagnesium iodide to produce $1 \ mol$ of methane,the moles of alcohol used is $0.005 \ mol$.
The molar mass of the alcohol is $\frac{\text{Given mass}}{\text{moles}} = \frac{0.44 \ g}{0.005 \ mol} = 88 \ g/mol$.
Among the options,$CH_{3}CH(OH)CH_{2}CH_{3}$ (butan$-2-$ol) has a molar mass of $74 \ g/mol$,while $C_{5}H_{12}O$ isomers have $88 \ g/mol$. The alcohol must be a primary alcohol to form an aldehyde with $PCC$ that gives a positive silver mirror test. The structure $CH_{3}CH(CH_{3})CH_{2}OH$ ($2$-methylpropan$-1-$ol) has a molar mass of $74 \ g/mol$. Re-evaluating the molar mass $88 \ g/mol$ corresponds to $C_{5}H_{12}O$. The alcohol $CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}OH$ or similar isomers fit. However,based on the requirement of forming an aldehyde that gives a silver mirror test,the alcohol must be primary. The correct structure is $CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}OH$ or similar. Given the options provided,the calculation leads to $88 \ g/mol$.