Preparation of alcohol Questions in English

(N/A) From Carboxylic Acids: Carboxylic acids are reduced to primary alcohols in excellent yields by the use of a strong reducing agent,lithium aluminium hydride $(LiAlH_{4})$.
Since $LiAlH_{4}$ is an expensive reagent,it is used only for preparing special chemicals.
$(b)$ From Esters: On an industrial scale,carboxylic acids are first converted to esters,which are then reduced to alcohols by hydrogenation in the presence of a catalyst.
$RCOOH + R'OH$ $\xrightarrow{H^+} RCOOR'$ $\xrightarrow{H_2, \text{catalyst}} RCH_2OH + R'OH$

(N/A) The reaction of carbonyl compounds $(C=O)$ with Grignard reagents $(RMgX)$ involves a nucleophilic addition to form an intermediate adduct,which upon hydrolysis yields an alcohol.
$(i)$ Formaldehyde $(HCHO)$ reacts with a Grignard reagent to form a primary $(1^{\circ})$ alcohol.
$(ii)$ Other aldehydes $(RCHO)$ react with Grignard reagents to form secondary $(2^{\circ})$ alcohols.
$(iii)$ Ketones $(RCOR')$ react with Grignard reagents to form tertiary $(3^{\circ})$ alcohols.

(N/A) Grignard reagents $(RMgX)$ react with methanal $(HCHO)$ to form primary alcohols with one additional carbon atom.
$(i)$ To prepare $C_6H_5CH_2OH$,the Grignard reagent required is phenylmagnesium bromide $(C_6H_5MgBr)$.
$C_6H_5MgBr + HCHO$ $\rightarrow C_6H_5CH_2OMgBr$ $\xrightarrow{H_3O^+} C_6H_5CH_2OH + Mg(OH)Br$
$(ii)$ To prepare $CH_3CH_2CH(CH_3)CH_2OH$,the Grignard reagent required is sec-butylmagnesium bromide $(CH_3CH_2CH(CH_3)MgBr)$.
$CH_3CH_2CH(CH_3)MgBr + HCHO$ $\rightarrow CH_3CH_2CH(CH_3)CH_2OMgBr$ $\xrightarrow{H_3O^+} CH_3CH_2CH(CH_3)CH_2OH + Mg(OH)Br$

(N/A) The acid-catalyzed hydration of alkenes involves the addition of water $(H_{2}O)$ across the double bond in the presence of an acid catalyst like $H_{2}SO_{4}$.
The mechanism proceeds as follows:
$1$. Protonation of the alkene to form a carbocation intermediate.
$2$. Nucleophilic attack by water on the carbocation.
$3$. Deprotonation to yield the alcohol.
This reaction follows the Markovnikov rule,where the hydroxyl group $(-OH)$ attaches to the more substituted carbon atom. The general reaction is:
$R-CH=CH_{2} + H_{2}O \xrightarrow{H^{+}} R-CH(OH)-CH_{3}$

(N/A) The mechanism of acid-catalyzed hydration of an alkene to form an alcohol proceeds in three steps:
Step-$1$: Protonation of alkene to form a carbocation by electrophilic attack of hydronium ion $(H_{3}O^{+})$.
$CH_{3}-CH=CH_{2} + H_{3}O^{+} \xrightarrow{\text{slow}} CH_{3}-CH^{+}-CH_{3} + H_{2}O$
Step-$2$: Nucleophilic attack of water on the carbocation.
$CH_{3}-CH^{+}-CH_{3} + H_{2}O \xrightarrow{\text{fast}} CH_{3}-CH(OH_{2}^{+})-CH_{3}$
Step-$3$: Deprotonation to form an alcohol.
$CH_{3}-CH(OH_{2}^{+})-CH_{3} + H_{2}O \xrightarrow{\text{fast}} CH_{3}-CH(OH)-CH_{3} + H_{3}O^{+}$

(N/A) Carboxylic acids are reduced to primary alcohols by lithium aluminium hydride $(LiAlH_{4})$ or more selectively with diborane $(B_{2}H_{6})$.
$RCOOH \xrightarrow[\text{(ii) } H_{3}O^{+}]{\text{(i) } LiAlH_{4} \text{ or } B_{2}H_{6}} RCH_{2}OH$
$(b)$ $(i)$ Diborane does not easily reduce functional groups such as esters,nitro,or halo groups.
$(ii)$ Sodium borohydride $(NaBH_{4})$ does not reduce the carboxyl group.
$(c)$ In this reaction,the reduction of $-COOH$ to $-CH_{2}OH$ is a partial reduction,whereas reduction by ($HI$ + Red $P$) is a complete reduction which yields an alkane.

(A) Ozonolysis of an alkene typically results in the formation of aldehydes,ketones,or carboxylic acids,depending on the structure of the alkene,not alcohols. Therefore,it is not a method for preparing alcohols.

(B) The reaction of a Grignard reagent $(RMgBr)$ with an aldehyde $(R'CHO)$ followed by acid hydrolysis $(H_3O^+)$ yields a secondary alcohol.
The reaction is:
$CH_3CH_2CHO + RMgBr \rightarrow CH_3CH_2CH(OMgBr)R$
$CH_3CH_2CH(OMgBr)R + H_3O^+ \rightarrow CH_3CH_2CH(OH)R + Mg(OH)Br$
Comparing the product $CH_3CH_2CH(OH)CH_2CH_3$ with the general formula $CH_3CH_2CH(OH)R$,we can see that $R$ must be an ethyl group,$CH_3CH_2-$.
Therefore,the correct option is $B$.

(C) Methanal $(HCHO)$ reacts with Grignard reagent $(RMgX)$ to form primary alcohols $(R-CH_2-OH)$.
Propan-$2$-$ol$ is a secondary alcohol,which is formed by the reaction of Grignard reagent with aldehydes other than methanal (e.g.,ethanal).
Therefore,propan-$2$-$ol$ cannot be prepared by the action of Grignard reagent on methanal.

(A) Acid-catalysed hydration of alkenes follows Markovnikov's rule,where the hydroxyl group $(-OH)$ attaches to the more substituted carbon atom.
$CH_2=CH_2 + H_2O \xrightarrow{H^+} CH_3CH_2OH$ (Ethanol).
$CH_3CH=CH_2 + H_2O \xrightarrow{H^+} CH_3CH(OH)CH_3$ (Propan$-2-$ol).
$CH_3CH_2CH=CH_2 + H_2O \xrightarrow{H^+} CH_3CH_2CH(OH)CH_3$ (Butan$-2-$ol).
Methanol $(CH_3OH)$ cannot be prepared by this method as it requires at least two carbon atoms.
Propan$-1-$ol and Butan$-1-$ol are primary alcohols that cannot be formed by direct hydration of simple alkenes due to Markovnikov's rule.
Therefore,Ethanol is the correct answer.

(C) Acid catalyzed hydration of alkenes follows Markovnikov's rule,where the hydroxyl group $(-OH)$ attaches to the more substituted carbon atom.
$Propan-1-ol$ $(CH_3CH_2CH_2OH)$ cannot be prepared by this method because the hydration of $propene$ $(CH_3CH=CH_2)$ yields $propan-2-ol$ as the major product according to Markovnikov's rule.
Therefore,$propan-1-ol$ is the correct answer.

(D) Hydroboration-oxidation of $Isobutylene$ $(CH_2=C(CH_3)_2)$ involves the addition of $H_2O$ across the double bond in an anti-$Markovnikov$ fashion.
In this reaction,the $OH$ group attaches to the less substituted carbon atom.
The reaction proceeds as follows:
$CH_2=C(CH_3)_2 + (H-BH_2)_2 \xrightarrow{H_2O_2, OH^-} (CH_3)_2CH-CH_2OH$
The product formed is isobutyl alcohol ($2-$methylpropan$-1-$ol).

(B) To prepare $3-$methylpentan$-3-$ol,we need a Grignard reagent to attack an ester. The structure of $3-$methylpentan$-3-$ol is $CH_3CH_2-C(OH)(CH_3)-CH_2CH_3$.
This tertiary alcohol can be formed by the reaction of an ester with two equivalents of a Grignard reagent.
Specifically,reacting ethyl propanoate $(CH_3CH_2COOC_2H_5)$ with two equivalents of methyl magnesium bromide $(CH_3MgBr)$ yields $3-$methylpentan$-3-$ol.
The reaction proceeds as follows:
$CH_3CH_2COOC_2H_5 + CH_3MgBr \rightarrow CH_3CH_2COCH_3 + C_2H_5OMgBr$
$CH_3CH_2COCH_3 + CH_3MgBr \rightarrow CH_3CH_2C(OMgBr)(CH_3)_2$
After hydrolysis: $CH_3CH_2C(OH)(CH_3)_2$ (which is $2-$methylbutan$-2-$ol). Wait,let us re-evaluate.
To get $3-$methylpentan$-3-$ol $(CH_3CH_2-C(OH)(CH_3)-CH_2CH_3)$,we need to react ethyl ethanoate $(CH_3COOC_2H_5)$ with two equivalents of ethyl magnesium bromide $(C_2H_5MgBr)$.
$CH_3COOC_2H_5 + 2C_2H_5MgBr \rightarrow CH_3C(OH)(C_2H_5)_2$.
Thus,the correct option is $B$.

(C) Let us analyze each reaction:
$A)$ Reaction of acetone $(CH_3COCH_3)$ with $CH_3MgI$ followed by hydrolysis gives $2-$methylpropan$-2-$ol.
$B)$ Hydroboration-oxidation of propene gives propan$-1-$ol.
$C)$ Acid-catalyzed hydration of propene follows Markovnikov's rule to give propan$-2-$ol.
$D)$ Reaction of formaldehyde $(HCHO)$ with $C_2H_5MgBr$ followed by hydrolysis gives propan$-1-$ol.
Therefore,the correct reaction is $C$.

(A) The preparation of $1$-ethyl cyclohexanol involves the acid-catalyzed hydration of ethylidene cyclohexane.
$1$. Ethylidene cyclohexane $(C_8H_{14})$ reacts with $H_3O^+$ (acid catalyst).
$2$. The double bond is protonated to form a stable tertiary carbocation at the $1$-position of the cyclohexane ring.
$3$. Water then attacks the carbocation,followed by deprotonation to yield $1$-ethyl cyclohexanol.
Thus,the correct reactants are ethylidene cyclohexane and $H_3O^+$.

(C) Ethyl magnesium bromide $(CH_3CH_2MgBr)$ is a Grignard reagent.
It reacts with acetone $(CH_3COCH_3)$ to form an addition product $X$ $(CH_3CH_2-C(OMgBr)(CH_3)_2)$.
Upon hydrolysis,this addition product $X$ yields $2-$methylbutan$-2-$ol $(CH_3CH_2-C(OH)(CH_3)_2)$.

(B) For the first reaction:
$CH_2O$ (formaldehyde) reacts with a Grignard reagent $(RMgBr)$ followed by hydrolysis to form a primary alcohol. The product is $CH_3CH_2CH_2CH_2OH$ (butan$-1-$ol). The Grignard reagent must be $CH_3CH_2CH_2MgBr$ (propylmagnesium bromide).
For the second reaction:
$Y$ reacts with $C_2H_5MgBr$ followed by hydrolysis to form $CH_3CH_2C(CH_3)_2OH$ ($2$-methylbutan$-2-$ol). This is a tertiary alcohol. The reaction of a ketone with a Grignard reagent produces a tertiary alcohol. Comparing the structure,$Y$ must be $CH_3COCH_3$ (acetone or propanone).
Thus,$X = CH_3CH_2CH_2MgBr$ and $Y = CH_3COCH_3$.

(B) The hydration of $2-$methyl$-2-$butene $(CH_3-C(CH_3)=CH-CH_3)$ follows Markovnikov's rule to form $2-$methyl$-2-$butanol $(X)$,which is a tertiary alcohol $(CH_3-C(OH)(CH_3)-CH_2-CH_3)$.
An isomer of $X$ is $3-$methyl$-2-$butanol $(CH_3-CH(OH)-CH(CH_3)_2)$,which is a secondary alcohol.
This isomer can be prepared by the reaction of acetaldehyde $(CH_3CHO)$ with isopropylmagnesium bromide $((CH_3)_2CHMgBr)$ followed by hydrolysis,as this is a Grignard reaction forming a secondary alcohol.
Thus,option $(B)$ is correct.

(A) $1$. Toluene reacts with $Br_2$ in the presence of $Fe$ (dark) via electrophilic aromatic substitution to form a mixture of $o$-bromotoluene $(P)$ and $p$-bromotoluene $(Q)$.
$2$. These isomers react with $Mg$ in dry ether to form the corresponding Grignard reagents: $o$-tolylmagnesium bromide and $p$-tolylmagnesium bromide.
$3$. These Grignard reagents react with formaldehyde $(HCHO)$ followed by acidic hydrolysis $(H_2O)$ to form primary alcohols.
$4$. The final products $R$ and $S$ are $o$-methylbenzyl alcohol and $p$-methylbenzyl alcohol,respectively.

(B) The target molecule is $3, 3-$dimethyl$-2-$butanol,which has the structure $(CH_3)_3C-CH(OH)-CH_3$.
$A$. Reaction of $3, 3-$dimethylbutanal with $MeMgBr$ followed by $H_3O^+$ gives $3, 3-$dimethyl$-2-$butanol. This is a valid preparation.
$B$. Acid-catalyzed hydration of $3, 3-$dimethyl$-1-$butene involves a carbocation rearrangement (ethyl shift) to form a more stable carbocation,leading to $2, 3-$dimethyl$-2-$butanol,not the target alcohol. Thus,$B$ cannot prepare it.
$C$. Ozonolysis of $2, 3, 4, 4-$tetramethyl$-2-$pentene followed by reduction with $NaBH_4$ yields $3, 3-$dimethyl$-2-$butanol and acetone. This is a valid preparation.
$D$. Reduction of $3, 3-$dimethyl$-2-$butanone with $LiAlH_4$ gives $3, 3-$dimethyl$-2-$butanol. This is a valid preparation.
$E$. Hydration of $3, 3-$dimethyl$-1-$butyne with $Hg^{2+}/H^+$ gives $3, 3-$dimethyl$-2-$butanone,not the target alcohol. Thus,$E$ cannot prepare it.
Therefore,$B$ and $E$ cannot be used to prepare the target alcohol.