1 methyl ethyne reacted with i(HBr,) ii(Mg(0), THF, acetone)

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The reaction of 1-methyl ethyne with HBr results in the formation of a bromoalkene due to the addition of Br across the triple bond. The Grignard reagent, generated from Mg(0) in THF, reacts with acetone to produce a tertiary alcohol. The final product is identified as 2-pentane-4-ol. Confirmation of this product is sought, indicating a need for peer validation. The discussion highlights the importance of understanding reaction mechanisms in organic chemistry.
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Homework Statement



I get that the Br will attach to the triple bond creating a double bond and a Br at the end. I get the gringard reagent but what will the acetone do?

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The Attempt at a Solution

 
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no one wants to respond :(
 
I believe I got it it is 2 pentane-4-ol. If anyone would like to confirm that it would be great.
 
Thread 'Confusion regarding a chemical kinetics problem'

Thread 'Confusion regarding a chemical kinetics problem'

TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
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