Question

Dalingalbabubnammatongmamalproducthahertamination Suprtytuaromintenanatimperpegation and termination steps.ciniantepmontageddhewondary and tetary probots. Oplulamompahyaram dely showWORD, POFTPEG 2.0- Line २२

The following radical substitution reaction gave several products after termination . Supply thw arrows in the initiation, propagation and termination steps. Calculate the percentage of the secondary and tertiary products.

Answer 1

Initiation cl + c د Propagation H, CH3 -1, HCI CH3 + CH3 د به ما و ملی - مه و متن - ساوه : من ي -- سموه CH3 CH3 H + HC - CH: H₃C CH3 Termination ه -- مو بلي - من.. CH3 ت - ده CH, CH3 سه - :CHس اها CH3 CH3 و CI ,CH نسمه I + * CH3 H- CH3 H CH3

The 2-methylbutane reacts with chlorine gas at hgher temperature and forms several compounds as shown below CH3 H3C I CH3 Hac + H - OCH + CACH Hoc Cha CH₃ CH₃ Y CH+ Y CH, + CHą H2C сня сі 2-methylbutane 1-chloro-2- 2-chloro-2- 2-chloro-3- 1-chloro-3- methylbutane methylbutane methylbutane methylbutane 25% 29.17% 33.33% 12.5% The formation of 1-chloro-2-methylbutane from 2-methylbutane with chlorine occurs through primary free radical formation. Thus, the pecent yield calculation is given below Compounds Types of hydrogen #H's % Yield Types of Radical Relative rates Relative reaction CH3 Fraction yield 6 x 100 1-chloro-2- methylbutane 1° CH3 6 X 1 = 6 = H2C CH3 2-chloro-2- methylbutane 3º x 7 = 1 = CH31 H3C CH3 I CH? H3CH2 2° 2-chloro-3- methylbutane 2 4 = 8 = CH3 1-chloro-3- methylbutane CH3 3 X 1 = 3 = > H2C 24 The percent yield formation of primary, secondary and tertiary alkyl halides are given above.