Preparation of alcohol Questions in English

(A) The indirect hydration of alkenes involves the reaction with concentrated $H_2SO_4$ followed by hydrolysis.
$CH_2=CH_2 + H_2SO_4 \to CH_3-CH_2-HSO_4 \xrightarrow{H_2O, \Delta} CH_3-CH_2OH + H_2SO_4$
Since the addition of $H_2SO_4$ to alkenes follows Markownikoff's rule,the hydrogen atom attaches to the carbon with more hydrogen atoms,and the $-HSO_4$ group attaches to the more substituted carbon.
For ethene $(CH_2=CH_2)$,this results in the formation of ethanol $(CH_3-CH_2OH)$.
For higher alkenes,this method typically yields secondary or tertiary alcohols. Therefore,ethyl alcohol is the only primary alcohol that can be prepared by this method.

(C) The industrial preparation of ethyl alcohol from ethylene involves the hydration of ethylene in the presence of concentrated $H_2SO_4$.
Step $1$: Ethylene reacts with concentrated $H_2SO_4$ to form ethyl hydrogen sulfate: $CH_2 = CH_2 + H_2SO_4 \rightarrow CH_3 - CH_2 - HSO_4$.
Step $2$: The resulting ethyl hydrogen sulfate is then hydrolyzed with water to produce ethyl alcohol: $CH_3 - CH_2 - HSO_4 + H_2O \rightarrow CH_3CH_2 - OH + H_2SO_4$.

(C) The reaction of $CH_3MgI$ with ethylene oxide (oxirane) follows a nucleophilic ring-opening mechanism.
$CH_3MgI$ acts as a nucleophile $(CH_3^-)$ and attacks the less hindered carbon of the ethylene oxide ring.
The resulting alkoxide intermediate,$CH_3-CH_2-CH_2-OMgI$,upon acidic hydrolysis,yields a primary alcohol,$CH_3-CH_2-CH_2-OH$ (propan$-1-$ol).
Reaction: $CH_2-CH_2(O) + CH_3MgI$ $\rightarrow CH_3-CH_2-CH_2-OMgI$ $\xrightarrow{H_2O/H^+} CH_3-CH_2-CH_2-OH + Mg(OH)I$.

(B) The reaction of Grignard reagent $(RMgX)$ with formaldehyde $(HCHO)$ followed by hydrolysis yields a primary alcohol:
$RMgX + HCHO$ $\rightarrow RCH_2OMgX$ $\xrightarrow{H_3O^+} RCH_2OH + Mg(OH)X$
Reaction with other aldehydes like $CH_3CHO$ yields secondary alcohols,while reaction with ketones yields tertiary alcohols.
Reaction with $CO_2$ yields carboxylic acids,and reaction with $H_2O$ yields alkanes.

(C) The industrial preparation of methanol involves the catalytic hydrogenation of carbon monoxide.
Water gas $(CO + H_2)$ is mixed with additional hydrogen gas.
Under the conditions of $673 \ K$ and $300 \ \text{atm}$ pressure,in the presence of a $Cr_2O_3/ZnO$ catalyst,the reaction proceeds as follows:
$CO + 2H_2 \xrightarrow{Cr_2O_3/ZnO, 673 \ K, 300 \ \text{atm}} CH_3OH$
Therefore,the correct product is methanol $(CH_3OH)$.

(B) The Grignard reagent $(C_2H_5MgBr)$ acts as a nucleophile and attacks the less hindered carbon of the ethylene oxide (oxirane) ring.
This results in the opening of the ring to form an alkoxide intermediate: $C_2H_5-CH_2-CH_2-OMgBr$.
Subsequent hydrolysis with $H_2O$ converts the alkoxide into a primary alcohol: $C_2H_5-CH_2-CH_2-OH$ ($n$-butanol).
The overall reaction is: $C_2H_5MgBr + CH_2(O)CH_2 \xrightarrow{H_2O} C_2H_5-CH_2-CH_2-OH$.

(B) The reaction between a Grignard reagent $(C_2H_5MgI)$ and formaldehyde $(HCHO)$ proceeds as follows:
$HCHO + C_2H_5MgI \rightarrow C_2H_5-CH_2-OMgI$
Upon subsequent hydrolysis,the intermediate forms a primary alcohol:
$C_2H_5-CH_2-OMgI + H_2O \rightarrow C_2H_5-CH_2-OH + Mg(OH)I$
The final product is $C_3H_7OH$ (propan$-1-$ol).

(D) The reaction of a Grignard reagent $(RMgBr)$ with oxygen $(O_2)$ proceeds as follows:
$RMgBr + O_2 \rightarrow R-OMgBr$
This intermediate,$R-OMgBr$,upon subsequent hydrolysis,yields an alcohol $(ROH)$:
$R-OMgBr + H_2O \rightarrow ROH + Mg(OH)Br$
Therefore,the final product is an alcohol $(ROH)$.

(C) Ethanol $(C_2H_5OH)$ can be synthesized as a byproduct when an ethyl ester like ethyl acetate $(CH_3COOC_2H_5)$ reacts with a Grignard reagent $(CH_3MgI)$.
The reaction involves the nucleophilic attack of the Grignard reagent on the carbonyl carbon,leading to the displacement of the ethoxide group $(C_2H_5O^-)$.
Upon hydrolysis,this ethoxide group forms ethanol $(C_2H_5OH)$.
Reaction:
$CH_3COOC_2H_5 + CH_3MgI \rightarrow CH_3COCH_3 + C_2H_5OMgI$
$C_2H_5OMgI + H_2O \rightarrow C_2H_5OH + Mg(OH)I$

(D) Alcoholic fermentation is a biological process in which sugars such as glucose,fructose,and sucrose are converted into cellular energy and thereby produce ethanol and carbon dioxide as metabolic waste products.
This process is catalyzed by the enzyme complex $Zymase$,which is naturally found in $Yeast$ $(Saccharomyces \text{ } cerevisiae)$.
The chemical reaction is: $C_6H_{12}O_6 \xrightarrow{Zymase \text{ (from yeast)}} 2C_2H_5OH + 2CO_2$.

(C) The reaction of methyl magnesium iodide $(CH_3MgI)$ with acetone $(CH_3COCH_3)$ followed by acid hydrolysis yields tertiary butyl alcohol.
$CH_3COCH_3 + CH_3MgI \rightarrow (CH_3)_3COMgI$
$(CH_3)_3COMgI + H_2O \rightarrow (CH_3)_3COH + Mg(OH)I$
Thus,the correct option is $(C)$.

(C) The reaction of a Grignard reagent $(CH_3MgI)$ with formaldehyde $(HCHO)$ followed by acidic hydrolysis yields a primary alcohol.
The nucleophilic methyl group from the Grignard reagent attacks the electrophilic carbonyl carbon of formaldehyde to form an intermediate alkoxide:
$HCHO + CH_3MgI \to CH_3CH_2OMgI$
Subsequent hydrolysis of the intermediate results in the formation of ethanol:
$CH_3CH_2OMgI + H_2O \to CH_3CH_2OH + Mg(OH)I$
Thus,the final product is $C_2H_5OH$.

(A) The reaction of alkyl halide $(R-X)$ with aqueous $KOH$ undergoes nucleophilic substitution ($S_N1$ or $S_N2$) to form alcohol $(R-OH)$.
$R-X + KOH_{(aq)} \rightarrow R-OH + KX$
Alcoholic $KOH$ is used for dehydrohalogenation to form alkenes.

(D) The hydration of alkenes to form alcohols using $Al_2O_3$ as a catalyst is a surface-catalyzed reaction.
In heterogeneous catalysis,the reaction rate and efficiency are primarily dependent on the surface area of the catalyst available for the adsorption of reactants.
Therefore,the surface area of $Al_2O_3$ is the most effective factor for this transformation.

(C) Industrially,ethylene is converted to ethyl alcohol by the hydration of ethylene in the presence of concentrated $H_2SO_4$.
The reaction proceeds as follows:
$CH_2 = CH_2 + H_2SO_4 \to CH_3CH_2HSO_4$
$CH_3CH_2HSO_4 + H_2O \xrightarrow{\Delta} CH_3CH_2OH + H_2SO_4$

(C) Isopropyl alcohol $(CH_3CH(OH)CH_3)$ is a secondary alcohol. It can be prepared by the reaction of a Grignard reagent with an aldehyde other than formaldehyde. Specifically,the reaction of $CH_3MgI$ with acetaldehyde $(CH_3CHO)$ yields isopropyl alcohol after hydrolysis:
$CH_3CHO + CH_3MgI \rightarrow CH_3CH(OMgI)CH_3$
$CH_3CH(OMgI)CH_3 + H_2O/H^+ \rightarrow CH_3CH(OH)CH_3 + Mg(OH)I$
Therefore,the correct combination is $CH_3MgI$ and $CH_3CHO$.

(A) $1$. $CH_3-CH(OH)-CH_3 \xrightarrow{PBr_3} CH_3-CH(Br)-CH_3$ $(X)$ (Isopropyl bromide).
$2$. $CH_3-CH(Br)-CH_3 \xrightarrow{Mg, \text{dry ether}} CH_3-CH(MgBr)-CH_3$ $(Y)$ (Isopropyl magnesium bromide).
$3$. $CH_3-CH(MgBr)-CH_3 + C_2H_4O \rightarrow CH_3-CH(CH_3)-CH_2-CH_2-OMgBr$ $(Z)$ (Grignard reagent addition to epoxide).
$4$. $CH_3-CH(CH_3)-CH_2-CH_2-OMgBr \xrightarrow{H_2O, \Delta} CH_3-CH(CH_3)-CH_2-CH_2OH + Mg(OH)Br$ (Hydrolysis).
The final product is $3-methylbutan-1-ol$,which matches option $A$.

(A) The industrial production of methanol is carried out by the catalytic hydrogenation of carbon monoxide at high pressure and temperature in the presence of a catalyst mixture of $ZnO - Cr_2O_3$:
$CO(g) + 2H_2(g) \xrightarrow[200-300 \, atm]{ZnO - Cr_2O_3, 573-673 \, K} CH_3OH(l)$

(C) The reaction of phenylmagnesium bromide $(C_6H_5MgBr)$ with ethylene oxide $(C_2H_4O)$ followed by acid hydrolysis yields $2-\text{phenylethanol}$.
The reaction proceeds as follows:
$C_6H_5MgBr + C_2H_4O \rightarrow C_6H_5-CH_2-CH_2-OMgBr$
$C_6H_5-CH_2-CH_2-OMgBr + H^+/H_2O \rightarrow C_6H_5-CH_2-CH_2-OH + Mg(OH)Br$
Thus,the correct option is $C$.

(A) Formaldehyde $(HCHO)$ reacts with methyl magnesium bromide $(CH_3MgBr)$ to form an addition product,which upon hydrolysis yields ethanol $(C_2H_5OH)$.
$HCHO + CH_3MgBr \rightarrow CH_3CH_2OMgBr$
$CH_3CH_2OMgBr + H_2O \xrightarrow{H^+} CH_3CH_2OH + Mg(OH)Br$

(B) The reaction sequence shown is Oxymercuration-Demercuration.
In the first step,the alkene reacts with mercuric acetate $(Hg(CH_3COO)_2)$ in $THF$ and water to form an organomercury intermediate.
In the second step,reduction with $NaBH_4$ in a basic medium $(NaOH)$ replaces the mercury group with a hydrogen atom.
This process results in the Markovnikov addition of water across the double bond without carbocation rearrangement.
Therefore,$X$ is $Hg(CH_3COO)_2$.

(D) The reaction sequence is as follows:
$CH_3CH_2OH \xrightarrow{P + I_2} CH_3CH_2I (A)$
$CH_3CH_2I \xrightarrow{Mg, \text{ether}} CH_3CH_2MgI (B)$
$CH_3CH_2MgI + HCHO \rightarrow CH_3CH_2CH_2OMgI (C)$
$CH_3CH_2CH_2OMgI \xrightarrow{H_2O} CH_3CH_2CH_2OH (D) + Mg(OH)I$
Compound $D$ is $n-$propyl alcohol (propan$-1-$ol).

(C) The reaction involves the addition of a Grignard reagent to an aldehyde.
$1$. $CH_3-CH_2-CHO + CH_3MgI \xrightarrow{H_3O^+} CH_3-CH_2-CH(OH)-CH_3$.
$2$. This reaction directly yields the secondary alcohol $CH_3-CH_2-CH(CH_3)-OH$ from propanal $(CH_3-CH_2-CHO)$ using methylmagnesium iodide $(CH_3MgI)$ followed by acidic hydrolysis $(H_3O^+)$.
$3$. Therefore,the conversion is achieved in a single step using $CH_3MgI$ and $H_3O^+$.

(C) The target product is $CH_3-CH(CH_3)-CH(OH)-CH_3$,which is $3$-methylbutan-$2$-ol.
Option $A$: Acid-catalyzed hydration $(H_3O^+)$ of $3$-methylbut-$1$-ene $(CH_3-CH(CH_3)-CH=CH_2)$ involves the formation of a carbocation intermediate. The initial protonation gives a primary carbocation,which rearranges to a more stable tertiary carbocation $(CH_3-C^+(CH_3)-CH_2-CH_3)$. Attack by water leads to $2$-methylbutan-$2$-ol.
Option $B$: Hydroboration-oxidation follows anti-Markovnikov addition of water,yielding $3$-methylbutan-$1$-ol.
Option $C$: Oxymercuration-demercuration involves the Markovnikov addition of water without carbocation rearrangement. The $OH$ group adds to the more substituted carbon of the double bond. For $3$-methylbut-$1$-ene,the double bond is at the terminal position. However,the structure $CH_3-CH(CH_3)-CH(OH)-CH_3$ corresponds to the addition of $OH$ to the $C_2$ position. This reaction proceeds via a cyclic mercurinium ion,and water attacks the more substituted carbon $(C_2)$,resulting in $3$-methylbutan-$2$-ol as the major product.
Therefore,option $C$ is the correct reaction.

(C) $t$-butyl alcohol is $(CH_3)_3COH$,which is a tertiary alcohol.
Tertiary alcohols are synthesized by the reaction of a Grignard reagent with a ketone.
The reaction of methylmagnesium bromide $(CH_3MgBr)$ with acetone $(CH_3COCH_3)$ followed by acid hydrolysis yields $t$-butyl alcohol.
$CH_3MgBr + CH_3COCH_3$ $\xrightarrow{Et_2O} (CH_3)_3COMgBr$ $\xrightarrow{H_3O^{+}} (CH_3)_3COH + Mg(OH)Br$

(C) The reaction of $propan-1-ol$ with $SOCl_2$ in the presence of pyridine is a standard method for the preparation of chloroalkanes from alcohols.
This reaction proceeds via an $S_N2$ mechanism,which ensures the formation of $1$-chloropropane without rearrangement.
The reaction is: $CH_3CH_2CH_2OH + SOCl_2 \xrightarrow{\text{pyridine}} CH_3CH_2CH_2Cl + SO_2 + HCl$.
Option $A$ is incorrect because the addition of $HCl$ to propene follows Markovnikov's rule,yielding $2$-chloropropane.
Option $B$ is incorrect because treatment of propene with $Cl_2$ leads to addition across the double bond,and subsequent treatment with $aq. KOH$ would likely result in substitution or elimination,not specifically $1$-chloropropane.

(D) The hydration of an alkene follows $Markovnikov's$ rule,where the $OH$ group attaches to the more substituted carbon atom.
$Butan-1-ol$ can be prepared from $but-1-ene$.
$Butan-2-ol$ can be prepared from $but-1-ene$ or $but-2-ene$.
$3-Methylbutan-1-ol$ can be prepared from $3-methylbut-1-ene$.
$2,2-Dimethylpropan-1-ol$ (neopentyl alcohol) cannot be prepared by the direct hydration of any alkene because the corresponding alkene,$3,3-dimethylbut-1-ene$,would undergo rearrangement to form a more stable carbocation,leading to a different product (e.g.,$2,3-dimethylbutan-2-ol$).
Thus,the correct option is $(d)$.

(D) Step $1$: Reaction of isobutyl alcohol with $PBr_3$ gives isobutyl bromide: $(CH_3)_2CHCH_2OH + PBr_3 \rightarrow (CH_3)_2CHCH_2Br$.
Step $2$: Reaction with $Mg$ in dry ether forms the Grignard reagent: $(CH_3)_2CHCH_2Br + Mg \rightarrow (CH_3)_2CHCH_2MgBr$.
Step $3$: The Grignard reagent attacks ethylene oxide (epoxide) to form an alkoxide intermediate: $(CH_3)_2CHCH_2MgBr + C_2H_4O \rightarrow (CH_3)_2CHCH_2CH_2CH_2OMgBr$.
Step $4$: Acidic hydrolysis $(H_3O^+)$ yields the final alcohol product: $(CH_3)_2CHCH_2CH_2CH_2OH$ ($4$-methylpentan$-1-$ol).

(A) Oxymercuration-demercuration is a reaction that follows Markovnikov's rule for the addition of water across a double bond without rearrangement.
For allylbenzene $(C_6H_5-CH_2-CH=CH_2)$,the electrophilic mercuric species attacks the terminal carbon $(CH_2)$ to form a more stable carbocation-like intermediate,followed by the attack of water at the more substituted carbon $(CH)$.
This results in the formation of $1-$phenylpropan$-2-$ol as the major product.
The reaction is: $C_6H_5-CH_2-CH=CH_2 \xrightarrow[(ii) NaBH_4]{(i) Hg(OAc)_2} C_6H_5-CH_2-CH(OH)-CH_3$.

(A) The target molecule is $2\text{-phenylbutan-2-ol}$,which is a tertiary $(3^o)$ alcohol.
Option $A$: $HCHO + PhCH(CH_3)CH_2MgX$ gives a primary $(1^o)$ alcohol,$PhCH(CH_3)CH_2CH_2OH$.
Option $B$: $PhCOCH_2CH_3 + CH_3MgX$ gives $CH_3-CH_2-C(Ph)(OH)-CH_3$.
Option $C$: $PhCOCH_3 + CH_3CH_2MgX$ gives $CH_3-CH_2-C(Ph)(OH)-CH_3$.
Option $D$: $CH_3CH_2COCH_3 + PhMgX$ gives $CH_3-CH_2-C(Ph)(OH)-CH_3$.
Thus,option $A$ cannot prepare the target molecule.

(A) Step $1$: Dehydrohalogenation of $2-$chloro$-1-$phenylbutane with $EtOK/EtOH$ (a strong base) proceeds via the $E2$ mechanism to form the more stable conjugated alkene,$1-$phenylbut$-1-$ene $(X)$,as the major product.
Step $2$: The reaction of $1-$phenylbut$-1-$ene $(X)$ with $Hg(OAc)_2/H_2O$ followed by $NaBH_4$ is an Oxymercuration-Demercuration $(OMDM)$ reaction.
Step $3$: $OMDM$ follows the Markovnikov addition rule,where the $-OH$ group attaches to the more substituted carbon atom of the double bond.
Step $4$: In $1-$phenylbut$-1-$ene $(Ph-CH=CH-CH_2-CH_3)$,the carbon at position $1$ is attached to a phenyl group,making it more substituted than the carbon at position $2$. Therefore,the $-OH$ group adds to the $C1$ position,yielding $1-$phenylbutan$-1-$ol as the major product $Y$.

(C) The reaction sequence is as follows:
$1$. $CH_3CH_2OH + P + I_2 \rightarrow CH_3CH_2I$ ($A$ is ethyl iodide).
$2$. $CH_3CH_2I + Mg \xrightarrow{\text{Ether}} CH_3CH_2MgI$ ($B$ is ethylmagnesium iodide).
$3$. $CH_3CH_2MgI + HCHO \rightarrow CH_3CH_2CH_2OMgI$ ($C$ is an intermediate alkoxide).
$4$. $CH_3CH_2CH_2OMgI + H_2O \rightarrow CH_3CH_2CH_2OH + Mg(OH)I$ ($D$ is $n-$propyl alcohol).
Thus,the final product '$D$' is $n-$propyl alcohol.

(B) Oxymercuration-Demercuration $(OMDM)$ reaction involves the addition of $H_2O$ according to Markovnikov's rule without any carbocation rearrangement.
The reaction uses $Hg(OAc)_2/H_2O$ followed by reduction with $NaBH_4$.
In contrast,acid-catalyzed hydration (options $a$ and $d$) involves a carbocation intermediate which can undergo rearrangement to form a more stable carbocation.
Hydroboration-oxidation (option $c$) follows Anti-Markovnikov's rule.

(C) The conversion of propene $(CH_3-CH=CH_2)$ to propan$-1-$ol $(CH_3-CH_2-CH_2-OH)$ requires an anti-Markovnikov addition of water across the double bond.
This is achieved via the Hydroboration-Oxidation $(HBO)$ reaction.
In this process,propene reacts with diborane $(B_2H_6)$ to form an alkylborane intermediate,which is then oxidized by hydrogen peroxide $(H_2O_2)$ in the presence of an aqueous base $(OH^\Theta)$ to yield the primary alcohol.
Therefore,the correct reagent sequence is $B_2H_6$ followed by $H_2O_2/OH^\Theta$.

(B) $Grignard$ reagents $(RMgX)$ react with aldehydes (other than formaldehyde) to produce secondary $(2^o)$ alcohols after hydrolysis.
Acetaldehyde $(CH_3CHO)$ is an aldehyde that reacts with $RMgX$ to form $R-CH(OH)-CH_3$,which is a $2^o$ alcohol.
Reaction: $CH_3CHO + RMgX \xrightarrow{H_2O} CH_3-CH(OH)-R$.
Formaldehyde $(HCHO)$ gives a $1^o$ alcohol,and ketones like acetone $(CH_3COCH_3)$ give a $3^o$ alcohol.

(C) Let us analyze each reaction:
$(A)$ $CH_3-CH=CH_2 \xrightarrow{H_3O^{\oplus}}$ follows Markovnikov's rule to form $CH_3-CH(OH)-CH_3$,which is a secondary alcohol.
$(B)$ $CH_3-CH=O \xrightarrow[(2) \, H_2O]{(1) \, CH_3MgBr}$ forms $CH_3-CH(OH)-CH_3$,which is a secondary alcohol.
$(C)$ $CH_3-CH=CH_2 \xrightarrow[(2) \, H_2O_2/\overset{\Theta}{O}H(aq.)]{(1) \, BH_3, THF}$ is the hydroboration-oxidation reaction,which follows anti-Markovnikov's rule to form $CH_3-CH_2-CH_2OH$,which is a primary alcohol.
Therefore,the correct option is $C$.

(C) Grignard reagent $(RMgX)$ reacts with formaldehyde $(HCHO)$ to form a primary alcohol.
The general reaction is: $RMgX + HCHO \xrightarrow{H_3O^+} R-CH_2-OH$.
Given product is $CH_3-CH(CH_3)-CH_2-OH$.
By comparing $R-CH_2-OH$ with $CH_3-CH(CH_3)-CH_2-OH$,we get $R = CH_3-CH(CH_3)-$,which is an isopropyl group.

(A) The given reaction is an example of oxymercuration-demercuration of an alkene.
This reaction follows Markovnikov's rule for the addition of $H_2O$ across the double bond.
In the reaction $(CH_3)_3C-CH=CH_2$,the alkene is terminal.
According to Markovnikov's rule,the $OH^-$ group attaches to the more substituted carbon atom (the carbon atom with fewer hydrogen atoms).
Here,the carbon atom at the $CH$ position is more substituted than the terminal $CH_2$ group.
Therefore,the $OH$ group attaches to the $CH$ carbon,resulting in the product $(CH_3)_3C-CH(OH)CH_3$.

(B) The preparation of alkyl chlorides from alcohols involves the substitution of the $-OH$ group with a $Cl$ atom.
$1$. $HCl +$ Anhydrous $ZnCl_2$ (Lucas reagent) is used to convert alcohols to alkyl chlorides.
$2$. $PCl_5$ reacts with alcohols to form alkyl chlorides $(R-OH + PCl_5 \rightarrow R-Cl + POCl_3 + HCl)$.
$3$. $SOCl_2$ (Thionyl chloride) is the preferred reagent for preparing alkyl chlorides because the by-products ($SO_2$ and $HCl$) are gases,leaving the pure alkyl chloride.
$4$. $NaCl$ is an ionic compound and cannot directly substitute the $-OH$ group of an alcohol to form an alkyl chloride under standard conditions,as the $Cl^-$ ion is a poor nucleophile in this context and the $-OH$ group is a poor leaving group.
Therefore,$NaCl$ is not useful for this transformation.

(A) When an alkyl halide $(R-X)$ is treated with moist silver oxide $(Ag_2O + H_2O)$,it acts as a source of $AgOH$.
The reaction proceeds as follows:
$R-X + AgOH \rightarrow R-OH + AgX$.
The product formed is an alcohol (alkanol).

(C) The structure of $3-$ethylpentan$-3-$ol is $(CH_3CH_2)_3C-OH$.
This is a tertiary alcohol. Grignard reagents react with ketones to form tertiary alcohols.
To obtain $(CH_3CH_2)_3C-OH$,we need a ketone and a Grignard reagent that provide the three ethyl groups attached to the central carbon.
Specifically,the reaction between diethyl ketone (pentan$-3-$one) and ethylmagnesium bromide is:
$CH_3CH_2COCH_2CH_3 + CH_3CH_2MgBr \rightarrow (CH_3CH_2)_3C-OMgBr$
Followed by acid hydrolysis:
$(CH_3CH_2)_3C-OMgBr + H_2O \rightarrow (CH_3CH_2)_3C-OH + Mg(OH)Br$
Thus,the correct reagents are $CH_3CH_2MgBr$ and $CH_3CH_2COCH_2CH_3$.

(D) Step $1$: $CH_3CH_2OH + P + I_2 \rightarrow CH_3CH_2I$ ($A$ is ethyl iodide).
Step $2$: $CH_3CH_2I + Mg \xrightarrow{\text{ether}} CH_3CH_2MgI$ ($B$ is ethyl magnesium iodide,a Grignard reagent).
Step $3$: $CH_3CH_2MgI + HCHO \rightarrow CH_3CH_2CH_2OMgI$ ($C$ is the addition product).
Step $4$: $CH_3CH_2CH_2OMgI + H_2O \rightarrow CH_3CH_2CH_2OH + Mg(OH)I$ ($D$ is $n-$propyl alcohol or propan$-1-$ol).

(B) Reaction $(i)$ involves the hydrolysis of ethyl bromide with water,which is a slow reversible reaction and produces $HBr$,which can push the equilibrium backward.
Reaction $(ii)$ uses moist silver oxide $(Ag_2O + H_2O \longrightarrow 2AgOH)$. The $AgOH$ reacts with $CH_3CH_2Br$ to form $CH_3CH_2OH$ and $AgBr$. The precipitation of $AgBr$ drives the reaction to completion,making it a much more efficient and easier method for the preparation of ethanol.

(N/A) $CH_3CH_2CH_2CH_2OH$ $(Butan-1-ol)$
$(b)$ $CH_3CH(OH)CH_3$ $(Propan-2-ol)$
$(c)$ $CH_3C(CH_3)(OH)CH_3$ $(2-Methylpropan-2-ol)$

(N/A) $(i)$ The reaction of isobutylmagnesium bromide with methanal $(HCHO)$:
$HCHO + CH_3-CH(CH_3)-CH_2-MgBr \to CH_3-CH(CH_3)-CH_2-CH_2-OMgBr$
$CH_3-CH(CH_3)-CH_2-CH_2-OMgBr \xrightarrow{H_2O/H^+} CH_3-CH(CH_3)-CH_2-CH_2-OH + Mg(OH)Br$
Note: The structure provided in the question $(i)$ is $CH_3-CH(CH_3)-CH_2OH$ (isobutanol),which is prepared from isopropylmagnesium bromide:
$HCHO + (CH_3)_2CH-MgBr \to (CH_3)_2CH-CH_2-OMgBr \xrightarrow{H_2O/H^+} (CH_3)_2CH-CH_2-OH + Mg(OH)Br$
$(ii)$ The reaction of cyclohexylmagnesium bromide with methanal $(HCHO)$:
$HCHO + C_6H_{11}-MgBr \to C_6H_{11}-CH_2-OMgBr$
$C_6H_{11}-CH_2-OMgBr \xrightarrow{H_2O/H^+} C_6H_{11}-CH_2-OH + Mg(OH)Br$

(N/A) $(i)$ By acid-catalyzed hydration of styrene $(phenylethene)$,$1-$phenylethanol can be synthesized:
$C_6H_5CH=CH_2 + H_2O \xrightarrow{H^+} C_6H_5CH(OH)CH_3$
$(ii)$ When chloromethylcyclohexane is treated with aqueous $NaOH$,cyclohexylmethanol is obtained via an $S_{N}2$ reaction:
$C_6H_{11}CH_2Cl + NaOH(aq) \to C_6H_{11}CH_2OH + NaCl$
$(iii)$ When $1-$chloropentane is treated with aqueous $NaOH$,pentan$-1-$ol is produced:
$CH_3CH_2CH_2CH_2CH_2Cl + NaOH(aq) \to CH_3CH_2CH_2CH_2CH_2OH + NaCl$

(N/A) The given alcohols can be synthesized by the acid-catalyzed hydration of appropriate alkenes,which follows Markovnikov's rule.
$(i)$ $1$-methylcyclohexene + $H_2O \xrightarrow{H^+} 1$-methylcyclohexanol
$(ii)$ $3$-methylhex-$2$-ene + $H_2O \xrightarrow{H^+} 3$-methylhexan-$3$-ol
$(iii)$ $pent-2-ene + H_2O \xrightarrow{H^+} pentan-2-ol$
$(iv)$ $2-cyclohexylbut-2-ene + H_2O \xrightarrow{H^+} 2-cyclohexylbutan-2-ol$

(N/A) Preparation: Alkenes react with water $(HOH)$ in the presence of an acid catalyst (dilute $HCl$ or $H_2SO_4$) to form alcohols.
If the alkene is unsymmetrical,the addition follows 'Markovnikov's rule',meaning the $-OH$ group attaches to the more substituted carbon and the $-H$ atom attaches to the less substituted carbon of the double bond.
$(b)$ Mechanism: The acid-catalyzed hydration of alkenes is an electrophilic addition reaction occurring in three steps.
Step-$1$: Protonation of the alkene by an electrophile $H_3O^+$ (or $H^+$) to form a more stable carbocation. This is the slow and rate-determining step as the $\pi$-bond of $C=C$ breaks.
Step-$2$: Nucleophilic attack of water $(H_2\ddot{O}:)$ on the carbocation $(M)$ to form a protonated alcohol.
Step-$3$: Deprotonation to form the final alcohol product.

(N/A) Preparation: Aldehydes and ketones undergo reduction to their corresponding alcohols by the addition of hydrogen in the presence of catalysts. This process is known as catalytic hydrogenation of aldehydes and ketones.
$(b)$ Catalysts: Finely divided metals such as platinum $(Pt)$,palladium $(Pd)$,or nickel $(Ni)$ are generally used for this reduction. Sodium borohydride $(NaBH_4)$ or lithium aluminium hydride $(LiAlH_4)$ can also be used to convert aldehydes and ketones into alcohols.
$(c)$ General Reaction: Aldehydes yield primary alcohols,and ketones yield secondary alcohols.
$(d)$ Examples:
$CH_3CHO + H_2 \xrightarrow{Pd/Pt} CH_3CH_2OH$
$C_6H_5CHO + H_2 \xrightarrow{Pd/Pt} C_6H_5CH_2OH$
$CH_3COCH_3 + H_2 \xrightarrow{NaBH_4} CH_3CH(OH)CH_3$
$C_6H_5COCH_3 + H_2 \xrightarrow{LiAlH_4} C_6H_5CH(OH)CH_3$