Draw the structure of the major organic product isolated from the reaction of 1-hexyne with sodium amide in liquid ammonia follo
wed by 1-bromobutane.
2 answers:
Answer:
CH3(CH2)3CCH --> CH3(CH2)3CCCH2(CH2)2CH3 + NaBr
Explanation:
This reaction are used in the elongation of a terminal alkyne chain, it can be in 2 steps:
Step 1:
NaNH2 is in this reaction is used in the deprotonation of 1-hexyne to give acetylide ions( RCC-). The ion is an nucleophile because of the lone pair of electrons.
Step 2:
This acetylide ion is a basic ion (because 1-hexyne is a weak acid). The alky halide (1-bromobutane) dissociates to form a primary cation and Br- ion. This cation then attacks the acetylide ion, reacts to form carbon-carbon bonds as well as add to carbonyl.
Below is an attachment for the chemical reaction.
Answer:check explanation and attached file/picture
Explanation:
Alkynes are hydrocarbons and they are very acidic because of hybridization effect(the more acidic the s-character is, the more acidic it is going to be).
The reaction of 1-hexyne with sodium amide in liquid ammonia is a form of deprotonation 'reaction' to form acetylide. Due to the acidic nature of the terminal hydrogen atom, terminal alkynes do form metallic derivatives by the replacement of the terminal hydrogens. The equation of Reaction is given below.
C6H10 + NaNH2( in liquidNH3) ------------> C6H9Na + H-NH2.
The acetylide is a bae and a very good nucleophile.
The reaction is then followed by the addition of 1-bromobutane. This reaction is used for the production of longer chain alkynes. The equation of Reaction is attached in the picture.
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The rate constant is mathematically given as
K2=2.67sec^{-1}
<h3>What is the Arrhenius equation?</h3>The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:
Therefore
KT1= 0.0110^{-1}
T1= 21+273.15
T1= 294.15K
T2= 200
T2=200+273.15
T2= 473.15K
Ea= 35.5 Kj/Mol
Hence, in j/mol R Ea is
Ea=35.5*1000 j/mol R
K2/0.0110 =e^(5.492)
K2/0.0110 =242.74
K2= 242.74*0.0110
K2=2.67sec^{-1}
In conclusion, rate constant
K2=2.67sec^{-1}
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Answer : The concentration of is,
Explanation :
When we assume this reaction is driven to completion because of the large excess of one ion then we are assuming limiting reagent is and
is excess reagent.
First we have to calculate the moles of KSCN.
Moles of KSCN = Moles of = Moles of
=
Now we have to calculate the concentration of
Total volume of solution = (6.00 + 5.00 + 14.00) = 25.00 mL = 0.025 L
Thus, the concentration of is,