Identify the condition for resonance

Resonance is shown by compounds in which there is conjugation, i.e. adjacent atoms with:

• a $\pi$ bond next to another $\pi$ bond, or
• a lone pair next to a $\pi$ bond, or
• a charge next to a $\pi$ bond.

So we check whether each compound has such delocalization.

Option A: $CH_3CH_2OCH=CH_2$

Structure can be written as: $CH_3CH_2-O-CH=CH_2$

Here oxygen has a lone pair adjacent to the double bond: $-O-CH=CH_2$

Therefore, resonance is possible by donation of lone pair from oxygen into the adjacent $\pi$ system: $CH_3CH_2-O-CH=CH_2 \leftrightarrow CH_3CH_2-O^+=CH-CH_2^-$

So A exhibits resonance.

Option C: $CH_3CH_2CH_2CONH_2$

This is an amide: $CH_3CH_2CH_2-C(=O)-NH_2$

In amides, the lone pair on nitrogen is conjugated with the carbonyl group: $R-C(=O)-NH_2 \leftrightarrow R-C(-O^-)=NH_2^+$

Hence C exhibits resonance strongly.

Option D: $CH_3CH_2CH=CHCH_2NH_2$

Structure: $CH_3CH_2-CH=CH-CH_2NH_2$

The $NH_2$ group is attached to a $CH_2$ carbon, not directly to the double bond carbon.

So the arrangement is: $C=C-CH_2-NH_2$

Because of the intervening saturated $CH_2$ carbon, the lone pair on nitrogen is not adjacent to the double bond. Thus there is no conjugation and hence no resonance involving the given functional groups.

So D does not exhibit resonance.

Conclusion

The compound that does not exhibit resonance is: $\boxed{D}$

Comparison with stored answer

Stored correct answer = D.

My derived answer = D.

Therefore, they agree.