Bromine Hopping

$Product \space B : \boxed{3}-bromo-3-ethyl pentane \\ Product \space D : \boxed{2}-bromo-3-ethyl pentane \\ c_3 - c_2 = 3 - 2 = \boxed{1} \Leftarrow Answer$ Step 1 : Removal of ketone groups

Step 2 : Addition of Bromine onto most highly substituted carbon : $c_3$

Step 3 : Removal of Bromine molecule to form a double bond

Step 4: Anti Markovnikov addition of Bromine onto least substituted carbon in double bond : $c_2$

Firstly you have clemmensen reduction which reduces ketone to alkane .

One interseting thing note about this reduction is that it cant be used to produce alkane if it contains an alcoholic group as it might get substituted as in Acidic medium OH becomes an extremely good leaving group

Other Similar reaction is Wolff Kishner reduction albeit it is in basic medium (Reagent Is Hydrazine+Base+Heat)

Think for which type of Ketone it can't produce alkane we desire!!

Next you have Free radical bromination .

The ratio of rate of reaction of Br2 with Primary , Secondary and tertiary hydrogens is 1:82:1600 .

So Obviously the Tertiary hydrogen will get substituted .

The Deciding factor for choosing hydrogen for substitution has contribution from two things one by ratio of reactivity and second Number of hydrogen atoms of that type (Probabilistic factor)

You Get 3-Bromo-3-ethylpentane

Next you have Bimolecular Elimination using Alcoholic KOH that is free of any rearrangement as it proceeds through a transition state(E2 Reaction)

Next is simple and most common HBr+Organic Peroxide (Free radical reaction) and follows Anti-Markovnikov's Regioselectivity

You get 2-Bromo-3-ethylpentane

Hence c1-c2 = 3-2 = 1

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