1. Identify the type of bond in each option

All the given compounds are vinylic chlorides of the form $\mathrm{Cl-CH=CHX}$, where $X$ is different:

• A: $\mathrm{Cl-CH=CH-CH_3}$
• B: $\mathrm{Cl-CH=CH-NO_2}$
• C: $\mathrm{Cl-CH=CH-OCH_3}$
• D: $\mathrm{Cl-CH=CH_2}$

So the $\mathrm{C-Cl}$ bond is attached to an $sp^2$ carbon in all cases.

2. Principle: when is a bond shorter?

A bond becomes shorter when it acquires partial double bond character.

For vinylic halides, the lone pair on chlorine can interact with the adjacent $\pi$ system:

$\mathrm{:Cl-CH=CHX \leftrightarrow Cl^+=CH-CH^-X}$

The more this resonance form is stabilized, the greater the partial double bond character in the $\mathrm{C-Cl}$ bond, hence the shorter the bond.

3. Effect of substituent $X$ on resonance stabilization

We compare how well the substituent on the other alkene carbon stabilizes the negative charge in the resonance form:

$\mathrm{Cl^+=CH-CH^-X}$

• If $X$ is electron-withdrawing by resonance/induction, it stabilizes the negative charge on the adjacent carbon.
• This increases contribution of the resonance structure with $\mathrm{C=Cl}$ character.
• Hence the $\mathrm{C-Cl}$ bond becomes shorter.

Now evaluate each substituent:

(A) $X = CH_3$

$\mathrm{CH_3}$ is weakly electron-donating by hyperconjugation, so it does not stabilize negative charge well.

(D) $X = H$

No special stabilization.

(C) $X = OCH_3$

$\mathrm{OCH_3}$ has a $+M$ effect (electron-donating by resonance), so it tends to destabilize the negative charge in the resonance form. Thus it reduces $\mathrm{C=Cl}$ character.

(B) $X = NO_2$

$\mathrm{NO_2}$ is a strong $-M$ and $-I$ group. It strongly withdraws electron density and stabilizes the negative charge in the resonance form. Therefore, it gives the maximum partial double bond character to the $\mathrm{C-Cl}$ bond.

So among all options, B should have the shortest $\mathrm{C-Cl}$ bond.

4. Order of $C-Cl$ bond length

Based on the above reasoning, shortest bond corresponds to strongest electron-withdrawing substituent:

$\mathrm{NO_2 > H \approx CH_3 > OCH_3}$

Thus the shortest $\mathrm{C-Cl}$ bond is in:

$\boxed{\mathrm{Cl-CH=CH-NO_2}}$

5. Compare with stored correct answer

Stored correct answer: B

My derived answer: B

They match.