Among the compounds A and B with molecular formula $\mathrm{C}_{9} \mathrm{H}_{18} \mathrm{O}_{3}, \mathrm{~A}$ is having higher boiling point the $B$. The possible structures of $A$ and $B$ are:

(D) $\mathrm{A}=$ (see structure) ; $\mathrm{B}=$ (see structure)

(A) $\mathrm{A}=$ (see structure) ; $\mathrm{B}=$ (see structure)

(B) $\mathrm{A}=$ (see structure) ; $\mathrm{B}=$ (see structure)

(C) $\mathrm{A}=$ (see structure) ; $\mathrm{B}=$ (see structure)

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Chemical Bonding and Molecular Structure

Among the compounds A and B with molecular formula

C_{9} H_{18} O_{3}, A

is having higher boiling point the

B

. The possible structures of

A

and

B

are:

Answer: (D)

Both isomers share the formula

C X_{9} H X_{18} O X_{3}

, so the difference in boiling point comes from hydrogen bonding. The higher-boiling

A

must carry groups that donate hydrogen bonds between molecules, a triol with three

- OH

groups that bind the molecules together strongly; the lower-boiling

B

has its oxygens locked as

- OCH X_{3}

ether groups, which cannot act as hydrogen-bond donors. So

A

is the triol and

B

the trimethoxy compound.

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Single correct option