When tert-pentylbromide is treated with 80% ethanol, the following amounts of al

Question

When tert-pentylbromide is treated with 80% ethanol, the following amounts of alkene products are deteced on ananlysis:

                   CH3

                    |               C2H5OH (80%)

CH3CH2----C-----Br -------------------------->    CH3CH = C(C(CH3)2 + CH3CH2C=CH2

                    |                                                                                                       |

                    CH3                                                                                                CH3

                                                                                      I                                    II

                                                                                    (32%)                            (8%)

                                                                       +{ tert-Pentyl alcohol

                                                                         { tert- Pently ethyl ether

                                                                                  (60%)

Offer an explanation of why compound I is formed in far greater amount than the terminal alkene.

Explanation / Answer

Tertiary compounds are more likely to undergo elimination reaction and Double bond at more hindered carbon is more stable than double bond at less hindered carbon,so compound 1 is formed in greater amounts than terminal alkene.

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