Question

Compounds X, C9H19Br, and Y, C9H19Cl, undergo base-promoted E2 elimination to give the same single alkene product, Z. Catalytic hydrogenation of Z affords 2,3,3-trimethylhexane. In water X readily reacts to give a substitution product, C9H19OH, together with Z. Y does not react with water. Propose structures for X and Y. Do not use stereobonds in your answer. In cases where there is more than one possible structure for each molecule, just give one for each. Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner. Separate structures with + signs from the drop-down menu.

Answer 1

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 methyl or hydride shift.

-) If we have an E2 reaction we will need an H in anti position 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 leaving group near to carbon that has more possible removable hydrogens.  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 alkyl bromide structure. To do this we have to check the alcohol produced by the alkene. In the hydration of alkanes reaction we will have a carbocation 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 methyl shift also, to obtain the same alcohol (figure 3).

Finally, for the alkyl chloride, we only have 2 choices because to produce the alkane we have to put the leaving group 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 methyl shift), therefore we have to put in the other carbon.