The hydrocarbon with seven carbon atoms containing a neopentyl and a vinyl group is:

First we recall the meanings of the two special names given in the statement:

• A vinyl group is the fragment $$CH_2=CH-$$. In other words, the double bond must be at the very end of the molecule so that the terminal carbon is $$CH_2$$ and is directly attached to the rest of the skeleton.

• A neopentyl group is the fragment $$(CH_3)_3CCH_2-$$. The characteristic feature is a $$CH_2-$$ unit that is bonded to a quaternary carbon bearing three methyl groups.

The required hydrocarbon must therefore contain both of these pieces somewhere in its structure and, when all carbon atoms are counted, the total must be seven.

Now we test every option one by one.

Option A : 2,2-dimethyl-4-pentene

Writing the carbon chain explicitly, we have

$$CH_3C(CH_3)_2CH_2CH_2CH=CH_2$$

Counting carbons: the pentane parent contributes $$5$$ and the two methyl groups contribute $$2$$, so the total is indeed $$7$$. However the double bond is shown at position 4 on the parent chain, so the fragment at the end is $$CH_3CH=$$, not $$CH_2=CH-$$. Therefore the terminal vinyl group is missing. Hence this option cannot satisfy the condition.

Option B : 4,4-dimethylpent-1-ene

First we write the parent chain of five carbons and place the double bond at carbon 1 as demanded by “pent-1-ene”:

$$CH_2=CH-CH_2-C-CH_3$$

The carbon labelled $$C$$ (carbon 4 in the name) still needs the two methyl substituents written in the name. Adding them gives

$$CH_2=CH-CH_2-C(CH_3)_2-CH_3$$

Now count the carbons:

• Parent chain = $$5$$.
• Two extra methyl groups = $$2$$.
Total $$= 5 + 2 = 7$$.

Next we look for the required fragments:

1. The left end is $$CH_2=CH-$$, exactly the vinyl group.

2. If we disconnect the bond immediately to the right of the $$CH_2-$$ unit, what remains on the right is $$C(CH_3)_3$$ : a quaternary carbon bonded to three methyl groups. Thus the fragment $$CH_2-C(CH_3)_3$$ is present, which is precisely the neopentyl group.

Hence this single structure simultaneously contains a vinyl group, a neopentyl group, and has seven carbon atoms. So Option B satisfies every requirement.

Option C : isopropyl-2-butene

Convert the name into a structure:

$$CH_3CH=CHCH_3$$ is the but-2-ene parent, and an isopropyl group $$CH_3CH(CH_3)-$$ is attached. Joining them gives a total of $$4 + 3 = 7$$ carbons; however the double bond is internal, so the molecule ends in $$CH_3CH=$$ rather than $$CH_2=CH-$$. Therefore the vinyl criterion fails, and there is no neopentyl fragment either.

Option D : 2,2-dimethyl-3-pentene

Its chain is

$$CH_3C(CH_3)_2CH=CHCH_3$$

Again the double bond lies inside the chain, giving no terminal $$CH_2=CH-$$ unit. Consequently the vinyl group is absent, so this option is also ruled out.

Among the four possibilities only Option B, 4,4-dimethylpent-1-ene, fulfils all the conditions laid down in the question.

Hence, the correct answer is Option B.