A magnet sticks to an iron can and an iron rod. This is an intrinsic property. True or false?
1 answer:
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Answer:
Rate constant of the reaction is .
Explanation:
A + B + C → D + E
Let the balanced reaction be ;
aA + bB + cC → dD + eE
Expression of rate law of the reaction will be written as:
Rate(R) of the reaction in trail 1 ,when :
...[1]
Rate(R) of the reaction in trail 2 ,when :
...[2]
Rate(R) of the reaction in trail 3 ,when :
...[3]
Rate(R) of the reaction in trail 4 ,when :
...[4]
By [1] ÷ [2], we get value of c ;
c = 1
By [3] ÷ [4], we get value of b ;
b = 0
By [2] ÷ [3], we get value of a ;
a = 2
Rate law of reaction is :
Rate constant of the reaction = k
Answer:
See explanation below
Explanation:
The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.
Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.
For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)
For question 6b, same principle of 6a, when the hydrogen migrates to the 2nd carbocation to form a tertiary carbocation the methanol will promove an E1 reaction with the vecinal carbons and form two eliminations products. See picture 2 for mechanism of reaction.
