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Kitty [74]
4 years ago
7

A chiral amine A having the R configuration undergoes Hofmann elimination to form an alkene B as the major product. B is oxidati

vely cleaved with ozone, followed by CH3SCH3, to form CH2═O and CH3CH2CH2CHO. What are the structures of A and B? Indicate stereochemistry where appropriate.

Chemistry
1 answer:
Free_Kalibri [48]4 years ago
3 0

Answer:

The Structure of "B" is alkene.

Explanation:

The compound "A" having R- configuration  and undergoes Hofmann elimination to form an alkene.

The compound "B" on oxidatively cleaving with ozone followed by dimethyl sulfide forms CH_{2}=S and CH_{3}CH_{2}CH_{2}CHO

The structure of "A" and"B" is as follows.

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Suppose a student prepares a buffer that is intended to be pH 7.20. However, when the student tests its pH using a pH probe, the
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Solid lead acetate is slowly added to 75.0 mL of a 0.0492 M sodium sulfate solution. What is the concentration of lead ion requi
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Answer:

The concentration of lead ion required to just initiate precipitation is -2.37\times10^-^5 M

Explanation:

Lets calculate -:

Solubility equilibrium -: PbI_2(s) ⇄ Pb^2^+ (aq) + 2I^- (aq)

Solubility product of PbI_2 ,Q=[Pb^2^+]_i_n_i_t_i_a_l [I^-]^2_i_n_i_t_i_a_l =9.8\times10^-^9

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When the ionic product exceeds the solubility product , precipitation of salt takes place .

                                   Q_s_p\geq K_s_p

        [Pb^2^+]_i_n_i_t_i_a_l [I^-]^2_i_n_i_t_i_a_l \geq 9.8\times10^-^9

      [Pb^2^+]_i_n_i_t_i_a_l  [0.0492]^2 \geq 9.8\times10^-^9

                       [Pb^2^+]_i_n_i_t_i_a_l \geq \frac{9.8\times10^-^9}{[0.0492]^2}

                        [Pb^2^+]_i_n_i_t_i_a_l \geq \frac{9.8\times10^-^9}{2.42\times10^-^3}

                        [Pb^2^+]_i_n_i_t_i_a_l \geq 2.37\times10^-^5 M

Thus , PbI_2 will start precipitating when [Pb^2^+]_i_n_i_t_i_a_l   \geq 2.37\times10^-^5 M.    

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