Which of the following compounds will show retention in configuration on nucleophilic substitution by $$OH^-$$ ion?

We begin by recalling that the word retention means that the absolute configuration (R or S) of the stereogenic centre remains the same after the substitution has taken place.

There are three usual mechanistic possibilities for a nucleophilic substitution on a chiral carbon bearing a leaving group X:

1. In a one-step $$S_N2$$ process the nucleophile approaches from the side opposite to the leaving group, causing a single inversion of configuration.
2. In a two-step $$S_N1$$ process the leaving group departs first, giving a planar carbocation. The nucleophile can then attack from either side, giving a racemic mixture (50 % retention + 50 % inversion in the ideal case).
3. In a mechanism involving neighbouring-group participation the atom of a neighbouring group with a lone pair (often another halogen) first displaces the leaving group intramolecularly. This produces a three-membered cyclic ion. The nucleophile then opens this ring from the back side. Each of the two intramolecular steps proceeds with inversion, so the overall result is double inversion => net retention.

With these facts we examine every option.

Option A is $$CH_3-\underset{\displaystyle *}{C(Br)(H)}-C_6H_{13}$$. Here the stereogenic carbon is directly bonded to Br, H, CH3 and C6H13. There is no neighbouring atom with a lone pair capable of participating. Hence the reaction will proceed either by $$S_N2$$ (one inversion) or, less likely, by $$S_N1$$ (racemisation). Retention is not expected.

Option B is $$CH_3-\underset{\displaystyle *}{CH(Br)}-C_6H_5$$ (benzyl bromide). The benzyl cation derived from it is highly stabilised, so an $$S_N1$$ pathway is favoured, producing racemisation. Again, no net retention.

Option C is $$CH_3-\underset{\displaystyle *}{CH(Br)}-CH_3$$ (sec-butyl bromide). Being a simple secondary bromide with no assisting group, it reacts mainly by $$S_N2$$ with a single inversion. Net retention is impossible.

Option D is $$CH_3-\underset{\displaystyle *}{CH(C_2H_5)}-CH_2Br$$. Here the stereogenic carbon is α to a $$CH_2Br$$ group. The bromine atom on the neighbouring carbon possesses lone pairs that can assist. Let us write every algebraic step.

Step 1: Intramolecular displacement (first inversion).
The lone pair on the neighbouring bromine attacks the stereogenic centre while the leaving bromide departs:

$$\text{(R or S)-CH}_3-\overset{*}{CH(C_2H_5)}-CH_2Br \;\xrightarrow{\text{l.p. on Br}}\; \bigl[\text{bicyclic bromonium ion}\bigr]^+ + Br^-$$

This internal attack must come from the back side of the C-Br bond that is breaking, so the stereogenic centre undergoes one inversion.

Step 2: Nucleophilic attack by $$OH^-$$ (second inversion).
The $$OH^-$$ ion now opens the three-membered bromonium ring from the side opposite to the bridging bromine:

$$\bigl[\text{bromonium ion}\bigr]^+ + OH^- \;\longrightarrow\; CH_3-\overset{*}{CH(C_2H_5)}-CH_2OH + Br^-$$

This ring opening is also a backside attack, so a second inversion occurs. Performing two consecutive inversions brings the configuration back to its original sense, i.e. overall retention.

Therefore only Option D provides the necessary neighbouring-group participation that gives double inversion and hence net retention.

Hence, the correct answer is Option 4.