Consider the following three-step reaction pathway.
Step 1: (CH3)3AUPH3 → (CH3)3Au + PH3 (fast)
Step 2: (CH3)3Au → (CH3)Au + C2H6 (slow)
Step 3: (CH3)Au + PH3 – (CH3)AuPH3 (fast)
Which is the correct list of reaction intermediates from the reaction pathway above?
A.) (CH3)3AuPH3, (CH3)3Au, and (CH3)Au
B.) PH3 and (CH3)Au
C.) (CH3)3AUPH3 and C2H6
D.) PH3, (CH3)3Au, and (CH3)Au

The answer is D.

Explanation:

We have to look at which products are created in reactions 1 and 2, that are used up in three.

If we look at Step 1, (CH3)3Au and PH3 are created. (CH3)3Au is used up as a reactant in step 2 and does not appear the final net reaction. Likewise, PH3 is used as a reactant in step 3 and does not appear in the net reaction.

If we look at Step 2, (CH3)Au and C2H6 are created. (CH3)Au is used up in the step 3 as a reactant and does not appear in the final reaction. C2H6 is never used as a reactant after being created in step 2, so it is not considered an intermediate since it appears in the final reaction.

You can check your work by looking at the net reaction:

(CH3)3AuPH3 → C2H6 + (CH3)AuPH3

(CH3)3Au, PH3, and (CH3)Au do not appear in the final reaction, and since they were created during the reaction, but were used up, it justifies that these are the three intermediates.

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