MCQ Questions for Class 12 Engineering Chemistry Alcohols,Phenols And Ethers Quiz 11

$CH_3-OH\xrightarrow{P+I_2}A\xrightarrow[alc\ KCN]{Heating}B\xrightarrow[Na/alcohol]{LiAlH_4}C\xrightarrow[NaNO_2\ + \ dil \ HCl] {HNO_2}D$

What is

$D$ ?

0%
$CH_3CHO$
0%
$CH_3COOH$
0%
$CH_3CH_2OH$
0%
$CH_3-O-CH_3$

Explanation

$CH_3-OH\xrightarrow{P+I_2}CH_3-I\xrightarrow[alc \ KCN]{Heating}CH_3-CN\xrightarrow[Na/ethyl \ alchohol]{LiAlH_4}CH_3-CH_2NH_2\xrightarrow[NaNO_2+dil \ HCl]{HNO_2}CH_3-CH_2OH$

Hence option C is correct.

Phenol is an organic compound even though it is soluble in water due to the presence of:

0%
ionic bonding
0%
covalent bonding
0%
hydrogen bonding
0%
coordinate bonding

Explanation

Phenol is an organic compound even though it is soluble in water due to the presence of hydrogen bonding. Phenol has hydroxyl group which is involved in the formation of hydrogen bonds with water. It increases the solubility. Hydrogen bonding is possible when

$H$ atom is attached to electronegative

$N,\ O$ or

$F$ atom.

Identify E:

0%
ethene
0%
ethane
0%
ethyne
0%
propyne

Product formed in the following reaction is/are :

0%
0%
0%
both (A) and (B)
0%
none is correct

Explanation

$Y$ is less soluble than

$\left(X\right)$ due to lack of symmetry of chiral carbon. This is reduced to

$-C{H}_{2}OH$ .

Under suitable condition temperature,pressure and catalyst,methanol can be obtain by:

0%
partial oxidation of hydrocarbons
0%
reacting $H_2$ with $CO$
0%
reacting $H_2$ with $CO_2$
0%
All the above

Explanation

Partial oxidation of hydrocarbons produces methanol.

$CO + 2H_2O \longrightarrow CH_3OH$

$CO_2 + H_2O \longrightarrow CH_3OH + H_2O$

All the three processes produces mathanol.

The major product of the reaction is:

0%
I
0%
II
0%
III
0%
IV

The major product of the following reaction is:

Explanation

The major product of the following reaction is represented by the option (C).

$\displaystyle Phenolic$

$\displaystyle -OH$ group is acidic in nature and on deprotonation gives

$\displaystyle phenoxide$ ion which then attacks

$\displaystyle methyl \: iodide$ . Thus net reaction is methylation of phenolic

$\displaystyle -OH$ group.

In the following sequence of reactions,

$CH_3Br\overset{KCN}{\rightarrow}A\overset{H_3O^+}{\rightarrow}B\overset{LiAlH_4}{\rightarrow}C$

The end product C is:

0%
Ethyl alcohol
0%
Acetaldehyde
0%
Acetone
0%
Methane

Explanation

The end product C is ethyl alcohol.

$\displaystyle CH_3Br \overset{KCN}{\rightarrow}\underset {A}{CH_3-CN}\overset{H_3O^+} {\rightarrow}\underset {B}{CH_3-COOH}\overset{LiAlH_4}{\rightarrow}\underset {C}{CH_3-CH_2-OH}$
Methyl bromide reacts with KCN to form acetonitrile. Br atom is replaced with CN group. Hydrolysis of cyano group gives carboxylic group. Reduction of carboxylic group gives hydroxyl group.

Compound

$'A'$ of molecular formula

$C_{4}H_{10}O$ on treatment with Lucas reagent at room temperature gives compound

$'B'$ . When compound

$'B'$ is heated with alcoholic

$KOH$ , it gives isobutene. Compound

$'A'$ and

$'B'$ are respectively:

0%
2-methyl-2-propanol and 2-methyl-2-chloropropane
0%
2-methyl-1-propanol and 1-chloro-2-methylpropane
0%
2-methyl-1-propanol and 2-methyl-2-chloropropane
0%
butan-2-ol and 2-chlorobutane
0%
none of the above

Explanation

Lucas test is used to distinguish between primary, secondary and tertiary alcohols, where

$OH$ is replaced by

$Cl$ . Since tertiary forms the stable carbocation, reacts readily with the reagent while primary react slowly upon heating. Therefore, out of possible alcohols 3 and 4, 3 is tertiary and reacts at room temperature. And alc. KOH upon heating with alkyl halides removes

$X\ and\ beta-H\ as\ HX$ resulting into alkene. Therefore compound 1 and 3 are A and B respectively.

Option A is correct.

Phenol is more acidic than:

0%
0%
0%
${ C }_{ 2 }{ H }_{ 2 }$
0%
Both (A) and (C)

Explanation

Methoxy group, due to

$+I$ -effect, increase electronic density on

$OH$ group, thus making it less acidic.

Thus,

$o$ -methoxy phenol and acetylene are less acidic than phenol

$p$ -nitrophenol is more acidic than phenol.

Phenol is more acidic than acetylene.

Option D is correct.

Isopropylbenzene is oxidized in the presence of air to compound

$'A'$ . When compound

$'A'$ is treated with dilute mineral acid, the aromatic product formed is:

0%
phenol
0%
benzene
0%
benzaldehyde
0%
acetophenone
0%
toluene

In the preparation of alkene from alcohol using

$Al_{2}O_{3}$ which is effective factor?

0%
Porosity of $Al_{2}O_{3}$
0%
Temperature
0%
Concentration
0%
Surface area of $Al_{2}O_{3}$

Explanation

Temperature is the effective factor for dehydration of alcohols by

${ Al }_{ 2 }{ O }_{ 3 }$ .

$R-C{ H }_{ 2 }-C{ H }_{ 2 }-OH\xrightarrow [ { 350 }^{ o }-{ 380 }^{ o }C ]{ { Al }_{ 2 }{ O }_{ 3 } }$

$R-CH=C{ H }_{ 2 }+{ H }_{ 2 }O$
At

${ 220 }^{ o }-{ 250 }^{ o }$ it forms ether.

Write the IUPAC name of

$\begin{matrix} & OH & \\ & | & \\ C{ H }_{ 3 }-C{ H }_{ 2 }- & C & -C{ H }_{ 2 }-C{ H }_{ 3 } \\ & | & \\ & C{ H }_{ 3 } & \end{matrix}$

0%
3-Methylpentan-3-ol
0%
3-Hydroxyhexane
0%
3-Hydroxy-3-methyl pentane
0%
All of the above

Explanation

The parent chain has 5 C atoms and an -OH group. So it is named pentanol. It has methyl substituent. The numbering of chain is shown below.

$\underset { 3-methyl-\quad pentan-\quad 3-ol }{ \begin{matrix} & OH & \\ & | & \\ \overset { 5 }{ C{ H }_{ 3 } } -\overset { 4 }{ C{ H }_{ 2 } } - & { C }^{ 3 } & -\overset { 2 }{ C{ H }_{ 2 } } -\overset { 1 }{ C{ H }_{ 3 } } \\ & | & \\ & C{ H }_{ 3 } & \end{matrix} }$

Hydroxy is used when

$-OH$ group is written in prefix. So, choice (b) and (c) are wrong.

Which of the following is the correct order regarding ease of dehydration of above compounds?

0%
I > III > II > IV
0%
IV > III > II > I
0%
IV > II > III > I
0%
III > IV > II > I

Choose the correct statement.

0%
$A$ is more volatile than $B$
0%
$B$ is more volatile than $A$
0%
$A$ is formed more rapidly at a higher temperature
0%
$A$ is formed in higher yield at a low temperature

Explanation

Boiling point of of

$A$ is lower than that of

$B$ . In

$A$ , there is intramolecular

$H-bonding$ and in

$B$ , there is intermolecular

$H-bonding$ ). Thus,

$A$ is more volatile than

$B$ .

In the given reaction, number of possible structure is:

0%
$1$
0%
$2$
0%
$5$
0%
$6$

Explanation

Therefore, in

$X$ order of stability of alkene is as follows:

$II> I> III$

All of these alkenes, with

${Br}_{2}/C{Cl}_{4}$ , produce additive product having molecular formula

${C}_{4}{H}_{6}{Br}_{2}$ . The possible products are

(i)

${ CH }_{ 3 }-\underset { |\\ Br }{ CH } -\underset { |\\ Br }{ CH } -{ CH }_{ 3 }$

(ii)

${ CH }_{ 3 }-\underset { |\\ Br }{CH}-\underset { |\\ Br }{ CH } -{ CH }_{ 3 }$

(iii)

$Br{ H }_{ 2 }C-{ CH }_{ 2 }-CHBr-{ CH }_{ 3 }\quad$

(iv)

${ CH }_{ 2 }Br-{ CH }_{ 2 }-{ CH }_{ 2 }-{ CH }_{ 2 }Br$

(v)

${ CH }_{ 3 }-{ CH }_{ 2 }-{ CH }Br-{ CH }_{ 2 }Br$

Synthesis of cyclohexane-1, 2-diol from cyclohexene may be accomplished in two way:

${MnO_4}^-$ dilute,

$OH^-$ at

$273 K$ .
peracid epoxidation followed by

$NaOH$ opening of the epoxide ring.
Which of the following statements/are correct :

0%
Method 1 and 2 give same product.
0%
Method 1 gives resolvable racemic mixture while 2 will give non resolvable achiral product.
0%
Method 1 gives non resolvable optivally inactive compound while method 2 gives resolvable racemic mixture.
0%
Product obtain in 1 and 2 will have diasteriomeric relationship.

The product of the reaction is:

Propan

$-2-ol\xrightarrow[250^0C]{Ai_2O_3}A\xrightarrow[HBr]{H_2O_2}B\xrightarrow{aq.KOH}$

What is the final product in the following reaction sequence?

0%
Propan-2-ol
0%
Propan-1-ol
0%
Ethane 1, 2 diol
0%
Propane 1, 2-diol

The relationship between the reactant and the final product is:

0%
position isomers
0%
identical compounds
0%
enantiomers
0%
none of the above

An alcohol

$(A)$ on dehydration gives

$(B)$ , which on Ozonolysis gives acetone and formaldehyde.

$(B)$ decolorises alkaline

$KMnO_{4}$ solution but

$(A)$ does not.

$(A)$ and

$(B)$ respectively are:

0%
$CH_{3}CH_{2}CH_{2}CH_{2}OH$ and $CH_{3}CH_{2}CH = CH_{2}$
0%
$CH_{3}CH_{2}-\overset {\overset {\displaystyle OH}{|}}{C}H-CH_{2}$ and $CH_{3} - CH = CH - CH_{3}$
0%
$(CH_{3})_{3}C - OH$ and $(CH_{3})_{2} C = CH_{2}$
0%
$CH_{3} - CH_{2} - \overset {\overset {\displaystyle OH}{|}}{C}H - CH_{3}$ and $(CH_{3})_{2}C = CH_{3}$

Ether is more volatile than alcohol having same molecular formula because of:

0%
dipolar character of ether
0%
alcohols having resonance structure
0%
intermolecular hydrogen bonding in ethers
0%
intermolecular hydrogen bonding in alcohols

Explanation

An ether is more volatile than an alcohol having same molecular formula because intermolecular hydrogen bonding in alcohols.
The

$\displaystyle -OH$ group of alcohol can form intermolecular hydrogen bonds wheres

$\displaystyle -O-$ group of ethers cannot form hydrogen bonds. Hydrogen bond formation is possible when H atom is attached to electronegative N, O or F atom. Intermolecular hydrogen bonds leads to molecular association.

Which of the following is not possible ?

0%
0%
0%
Both A and B
0%
None of these

The ease of dehydration in the following compounds is:

0%
$I > III > IV > II$
0%
$II > I > III > IV$
0%
$IV > I > III > II$
0%
$III > I > II > IV$

Acid catalysed dehydration of t-butanol is faster than that of n-butanol because:

0%
tertiary carbocation is more stable than primary carbocation
0%
primary carbocation is more stable than tertiary carbocation
0%
t-butanol has a higher boiling point
0%
- rearrangement takes place during dehydration of t-butanol.

Explanation

Alcohol

$+ H_2SO_4 \rightarrow$ Alkene

The dehydration reaction is an elimination reaction that goes via carbocation formation. Since tertiary carbocations are most stable due to + Inductive effect of alkyl groups, then is the secondary and least stable is primary.

Therefore the reactivity of alcohol is governed by the stability of carbocation that follows the order as

Tertiary > secondary > primary

Hence, acid catalysed dehydration of t-butanol is faster than that of n-butanol because tertiary carbocation formed by t-butanol is more stable than primary carbocation formed by n-butanol.

Rate of dehydration of the following in increasing order is:

0%
$I < II < III < IV$
0%
$II < III < IV < I$
0%
$I < III < IV < II$
0%
none of these

Explanation

Dehydration is the removal of water while having deprotonation from the adjacent carbon atom.

In the (IV) structure dehydration lead to formation of stable benzene and dehydration of (III) structure will have 2 alternate double in the ring which are in resonance.

Reactivity towards dehydration in (II) > (I) because (I) will destabilize the ring due to the formation of a triple bond in the ring.

Hence the Rate of dehydration in increasing order is-

$I<II<III<IV$ -Option A.

1516251_937685_ans_87b4cddef40945889d4d12f19bcdca00.png

The major product of acid catalysed dehydration of 2-methylcyclohexanol and butan -1-ol are respectively ?

0%
1-methylcyclohexence and but - 1-ene
0%
2-methylcyclohexence and but - 2-ene
0%
2-methylcyclohexence and butane
0%
1-methylcyclohexence and but -2-ene.

Explanation

Dehydration reaction of an alcohol can be represented as:

$CH-C-OH \xrightarrow {H^+} -C=C-$

Primary alcohol undergo dehydration via E2 mechanism and secondary and tertiary alcohol undergo dehydration by E1 mechanism via most stable carbocation formation through rearrangement.

For 2-methylcyclohexanol the dehydration product is 1-methylcyclohex-1-ene or 1-methylcyclohexene

and For Butan-1-ol the dehydration product is But-1-ene.

Which of the following is not a characteristic of alcohol ?

0%
They are lighter than water.
0%
Their boiling points rise fairly uniformly with rising molecular weight.
0%
Lower members are insoluble in water and organic solvents but the solubility regularly increases with molecular mass.
0%
Lower members have a pleasant smell and burning taste, higher members are colourless and tasteless.
0%
Higher members have a pleasant smell and burning taste, lower members are colourless and tasteless.

2.2 g of an alcohol (A) when treated with

$CH_3$ -MgI liberates 560 mL of

$CH_4$ at STP. Alcohol (A) on dehydration followed by ozonolysis gives ketone (B) along with (C). Oxime of ketone (B) contains 19.17% N. (A) on oxidation gives ketone (D) having same number of carbon atom .
Structure of (A) is ?

0%
$CH_3-\underset{CH_3}{\underset{|}C}H-CH_2-\underset{OH}{\underset{|}C}H-CH_3$
0%
$CH_3-\underset{CH_3}{\underset{|}C}H-CH-\underset{OH}{\underset{|}C}H-CH_3$
0%
$CH_3-CH_2-\underset{OH}{\underset{|}C}H-CH_2-CH_3$
0%
$CH_3-\overset{CH_3}{\overset{|}{\underset{CH_3}{\underset{|}C}}}-CH_2-OH$

Explanation

Given that:

$\underset {2.2g}{A(ROH)}+CH_3MgI \rightarrow \underset {560mL\ at\ STP}{CH_4}$

$A(ROH) \overset {(i)H^+(ii)Ozonolysis}{\rightarrow} B(ketone)+C$

$B+NH_2OH \rightarrow \underset {19.17N}{O(oxime)}$

$A(ROH) \overset {[O]}{\rightarrow} D(ketone$

560 mL of

$CH_4$ at STP:

number of moles of

$CH_4$ , n=560/22400=0.025 mol

1 mol of

$CH_4$ will be obtained by reaction of 1 mol of

$CH_3MgI$ with 1 mol of A

thus 0.025 mol of

$CH_4$ is obtained from 0.025 mol of A

if 0.025 mol of A=2.2g

mass of 1 mol of A=2.2/0.025=88g/mol

Molecular formula of an alcohol=

$C_nH_{2n+1}OH$

molar mass of alcohol=12n+2n+1+16+1

For A, molar mass=88,

thus, 12n+2n+1+16+1=88

and n=5

Thus A=

$C_5H_{11}OH$

Since A on dehydration will form an alkene which on ozonolysis gives a ketone. Therefore the alkene formed should have two alkyl groups attached on one of the doubly bonded carbon so as to get a ketone. Thus possible structure of alkenes are:

$CH_2={ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } -CH_2-C{ H }_{ 3 }$

$C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } =CH-C{ H }_{ 3 }$

Thus, possible structure of A are:

$\overset { \overset { OH }{ | } }{C}H_2-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } -CH_2-C{ H }_{ 3 }$ OR

$C{ H }_{ 3 }-\overset { \overset { OH }{ | } }{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } -CH_2-C{ H }_{ 3 }$

$C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } }H -\overset { \overset { OH }{ | } }{C}H-C{ H }_{ 3 }$

Thus possible B are:

$O={ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } -CH_2-C{ H }_{ 3 }$

$C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } =O$

and corresponding oxime are:

$HON={ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } -CH_2-C{ H }_{ 3 }$ =X

$C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } =NOH$ =Y

percent N in X is:(mass of N/molar mass of X)

$\times 100=\frac {14}{(12 \times 4+9+16+14)} \times 100=16.1$ %

percent N in Y is(:mass of N/molar mass of Y)

$\times 100=\frac {14}{(12 \times 3+7+16+14)} \times 100=19.18$ %

Therefore according to given data oxime obtained is

$C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } =NOH$

from ketone =B=

$C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } =O$

and possible alcohol are:

$C{ H }_{ 3 }-\overset { \overset { OH }{ | } }{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } -CH_2-C{ H }_{ 3 }$

$C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } }H -\overset { \overset { OH }{ | } }{C}H-C{ H }_{ 3 }$

Given that A gives ketone D on oxidation and with same number of carbons, thus it can only be a secondary alcohol as tertiary alcohol is resistant to oxidation. Thus A can only be:

$C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } }H -\overset { \overset { OH }{ | } }{C}H-C{ H }_{ 3 }$

overall reaction can be represented as:

$\underset {2.2g}{C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } }H -\overset { \overset { OH }{ | } }{C}H-C{ H }_{ 3 }}+CH_3MgI \rightarrow \underset {560mL\ at\ STP}{CH_4}$

$C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } }H -\overset { \overset { OH }{ | } }{C}H-C{ H }_{ 3 } \overset {(i)H^+(ii)Ozonolysis}{\rightarrow} C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } =O+O=CHCH_3$

$C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } =O+NH_2OH \rightarrow \underset {19.17N}{C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } =NOH}$

$C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } }H -\overset { \overset { OH }{ | } }{C}H-C{ H }_{ 3 } \overset {[O]}{\rightarrow}C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } }H -\overset { \overset { O }{ || } }{C}-C{ H }_{ 3 }$

Consider the following reaction sequence,

$CH_3CH(OH)CH_3 \xrightarrow[Heat]{Conc.H_2SO_4} X \xrightarrow[H_2O]{dil.H_2SO_4}Y$

X and Y in the reaction respectively are :

0%
$CH_3CH = CH_2, CH_3CH(OH)CH_3$
0%
$CH_3CH = CHCH_3, CH_3CH_2CH_2OH$
0%
$CH_3CH_2CH = CH_2,CH_3CH_2CH_2OH$
0%
$CH_3CH_2CH_2CH_3, CH_3CH(OH)CH_2CH_3$

Explanation

$CH_3 \underset {\underset {\displaystyle OH}{|}}{C}H-CH_3 \xrightarrow {Conc. H_2SO_4/Heat} \underset {(X)}{CH_3CH=CH_2} \xrightarrow {dil. H_2SO_4/H_2O} \underset {(Y)}{CH_3 \underset {\underset {\displaystyle OH}{|}}{C}H-CH_3}$

What happens when tertiary butyl alcohol is passed over heated copper at

$300^0$ C ?

0%
Secondary butyl alcohol is formed.
0%
2-Methylpropene is formed .
0%
1-Butene is formed .
0%
Butanal is formed.

Explanation

Tertiary alcohol when passed over

$Cu$ at

$300^{\circ}C$ is dehydrated to alkene.

$CH_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ \overset { \overset { C{ H }_{ 3 } }{ | } }{ C } } }-OH\ \overset { Cu,\ 300^{ 0 }C }{ \longrightarrow } CH_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ \overset { \overset { C{ H }_{ 2 } }{ || } }{ C } } }$

$t-butyl\ alcohol \overset { Cu,\ 300^{ 0 }C }{ \longrightarrow } 2-methyl\ propene$

Which of the following reactions will not yield phenol?

Major product obtained in the reaction is :

Which of the following is the correct order of boiling point?

0%
chloroethane $<$ ethane $<$ ethanol
0%
ethane $<$ chloroethane $<$ ethanol
0%
ethanol $<$ ethane $<$ chloroethane
0%
ethane $>$ chloroethane $>$ ethanol

Explanation

Ethanol being an alcohol will have the highest boiling point because hydrogen bonds exist between alcohol molecules. More heat is required to overcome the forces of attraction resulting from these hydrogen bonds.

Chloroethane has a higher boiling point than ethane because it has a

$Cl$ group. Chlorine is highly electronegative, making the chloroethane molecule polar in nature. The forces of attraction are higher among polar molecules and so, more heat is required to overcome the forces of attraction. Therefore, chloroethane, which is polar, has a higher boiling point than ethane, which is non-polar.

Thus the correct order of boiling point is:

Ethane < Chloroethane < Ethanol

Compare rate of dehydration of (i), (ii), and (iii) by conc.

$H_2SO_4$ .

0%
(i) > (iii) > (ii)
0%
(i) > (ii) > (iii)
0%
(ii) > (i) > (iii)
0%
(ii) > (iii) > (i)

Product (A) of the given reaction is?

Decreasing order of acidic strength of different (-OH) group is:

0%
$w > x > y > z$
0%
$w > z > x > y$
0%
$z > w > x > y$
0%
$z > x > w > y$

Explanation

$w$ will be the most acidic, as the anion (formed after the proton is removed) is stabilized by resonance in benzene ring.

$z$ will be the least acidic as the anion formed will experience a significant

$+I$ effect from the

$-CH_2$ group adjacent to it, thus destabilizing it.

$x$ will be comparatively more acidic than

$y$ as it experiences less

$+I$ effect to a less extent. Hence the order of acidic strength is:

$w>x>y>z$

A solution of

$Ph_{3}CCO_{2}H$ in conc.

$H_{2}SO_{4}$ gives

$(X)$ when poured into methanol

$X$ is :

0%
$Ph_3C-\overset{\overset{O}{||}}{C}-O-CH_3$
0%
$Ph_2CH-\overset{\overset{O}{||}}{C}-O=CH_3$
0%
$Ph_{3}C - OCH_{3}$
0%
$Ph_{3}C - CH_{3}$

Explanation

$Ph_3C-C-OOH+CH_3OH\xrightarrow []{conc.H_2SO_4}Ph_3C-OCH_3$

Mechanism:-

$Ph_3C^{\oplus}$ is a stable carbocation whicch breaks replaces

$H^+$ in

$CH_3OH$ to give

$Ph_3C-OCH_3$ ;

(Refer to Image)

1083184_993139_ans_1ceb370c872a4da2bdd3c0efb181df99.png

Based on above isomers answer the following.

Which isomers on dehydration give most stable alkene?

0%
a, b, g,
0%
d, e, f, g, h
0%
c,d, b, g
0%
None of these

Major product which you expect in the above reaction is :

Explanation

Iodide

$(I^-)$ from

$NaI$ undergoes an

$S_N2$ substitution and attacks the Carbon atom of

$C-OTs$ from behind. Hence, an inversion of stereochemistry occurs and we obtain the product given in Option C.
Hence, Option C is the correct answer.

1055163_989681_ans_e4ea9577deeb4ac4931dff5e7e96aeb5.png