#### Answer

$\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CBrCH}_{2} \mathrm{CH}_{3}$

#### Work Step by Step

See answer above.

Published by
Pearson Prentice Hal

ISBN 10:
0132064529

ISBN 13:
978-0-13206-452-1

$\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CBrCH}_{2} \mathrm{CH}_{3}$

See answer above.

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