Answer:
Here's what I get
Explanation:
You may have done a Williamson synthesis of guaifenesin by reacting guaiacol with 3-chloropropane-1,2-diol.
A. Mechanism
Step 1
NaOH converts guaiacol into a phenoxide ion.
Step 2
The phenoxide acts as the nucleophile in an SN2 reaction to displace the Cl from the alkyl halide.
B. Improve the yield
You probably carried out the reaction in ethanol solution — a polar protic solvent.
You might try doing the reaction in a polar aprotic solvent— perhaps DMSO.
A polar aprotic solvent does not hydrogen bond to nucleophiles, so they become stronger.
C. Another method of ether synthesis —dehydration of alcohols
Sulfuric acid catalyzes the conversion of primary alcohols to ethers.
This is also a nucleophilic displacement reaction.
Protonation of the OH converts it into a better leaving group.
Attack by a second molecule of alcohol forms the protonated ether.
A molecule of water then removes the proton.
Answer:
The six member ring and the position of the -OH group on the carbon (#4) identifies glucose from the -OH on C # 4 in a down projection in the Haworth structure). Fructose is recognized by having a five member ring and having six carbons, a hexose.
Answer:all
Explanation:
Because a reaction does not starts by itself unless some force is applied
Explanation:
Equation of the reaction:
Br2(l) + Cl2(g) --> 2BrCl(g)
The enthalpy change for this reaction will be equal to twice the standard enthalpy change of formation for bromine monochloride, BrCl.
The standard enthalpy change of formation for a compound,
ΔH°f, is the change in enthalpy when one mole of that compound is formed from its constituent elements in their standard state at a pressure of 1 atm.
This means that the standard enthalpy change of formation will correspond to the change in enthalpy associated with this reaction
1/2Br2(g) + 1/2Cl2(g) → BrCl(g)
Here, ΔH°rxn = ΔH°f
This means that the enthalpy change for this reaction will be twice the value of ΔH°f = 2 moles BrCl
Using Hess' law,
ΔH°f = total energy of reactant - total energy of product
= (1/2 * (+112) + 1/2 * (+121)) - 14.7
= 101.8 kJ/mol
ΔH°rxn = 101.8 kJ/mol.
Localized molecular orbitals are molecular orbitals which are concentrated in a limited spatial region of a molecule, for example a specific bond or a lone lake on a specific atom.
