Answer:
Compound X= 4-bromo-2,3,3-trimethylhexane
Compound Y= 5-chloro-2,3,3-trimethylhexane
Explanation:
The first step is set up the problem. That way we can obtain some clues. If we check figure 1 we can obtain some ideas:
-) If we have E2 reaction is not possible a <u>methyl or hydride shift</u>.
-) If we have an E2 reaction we will need an H in <u>anti position</u> to obtain the double bond. Therefore a double bond with the quaternary carbon (the carbon bonded to the 2 methyl groups).
The second step is to solve the alkene structure. We have to put the <u>leaving group</u> near to carbon that has more possible <u>removable hydrogens</u>. That's why the double bond is put it between carbons 5 and 4 of the alkane (Figure 2).
The third step is the structure of the <u>alkyl bromide</u> structure. To do this we have to check the alcohol produced by the alkene. In the <u>hydration of alkanes</u> reaction we will have a <u>carbocation</u> formation. Therefore we can have for the alkene proposed a methyl shift to obtain the most stable carbocation. With this in mind, we have to do the same for the Alkyl bromide that's why the Br is put it carbon 4 of the alkane. If we put the Br on this carbon we can have the chance of this <u>methyl shift</u> also, to obtain the same alcohol (figure 3).
Finally, for the <u>alkyl chloride</u>, we only have 2 choices because to produce the alkane we have to put the <u>leaving group</u> on one of the 2 carbons of the double bond. If we choose the same carbon on which we put the Br we can have the same behavior of the alkyl bromide (the <u>methyl shift</u>), therefore we have to put in the other carbon.
