How to replace multiple sub occurrences of string

I have one use case where I can have a text string which can contain anything. What I want to achieve is to replace a certain pattern within that given string.

Let's say I have given string as

:es1
es2
aaes1aa
:es3,
es1:
ees1,
ees1 

{
"es1 :

What I am trying to do is here is suppose I have to replace all es1 in this string but with one condition. It has to be either end or start with [, | ; | : | " | ' | \\ | \s]. :es1, es1,, es1: and so on are accepted but eees1sss is not.

I tried ([, | ; | : | " | ' | \\ | \s])(es1)([, | ; | : | " | ' | , | \s]) something like this but I don't think it's what I need.

Go program:

match := regexp.MustCompile(`([, | ; | : | " | ' | \\ | \s])(es1)([, | ; | : | " | ' | , | \s])`)
	test := `:es1
	es2
	aaes1aa
	:es3,
	es1:
	ees1,
	ees1 
	
	{
	"es1 :`
	
	fmt.Println(match.ReplaceAllString(test, "$1es4$3"))

output:

       es2
        aaes1aa
        :es3,
:
        ees1,
        ees1 

        {
         :

I was expecting my output to be more like

:es4
    es2
    aaes1aa
    :es3,
    es4:
    ees1,
    ees1 

    {
    "es4 :

the solution provided below is not well tested against all possibilities, but it seems to be working.

package main

import (
	"fmt"
	"regexp"
)

func main() {
	match := regexp.MustCompile(`([, | ; | : | " | ' | \\ | \s])es1|^es1([, | ; | : | " | ' | , | \s])`)
	test := `:es1
    es2
    aaes1aa
    :es3,
    es1:
    ees1,
    ees1 
    
    {
    "es1 :`

	fmt.Println(match.ReplaceAllString(test, "es4"))
}

https://play.golang.org/p/E8lb9vmM_Sa

You can use

package main

import (
	"fmt"
	"regexp"
)

func main() {
	match := regexp.MustCompile(`([,;:"'\\\s])es1\b|\bes1([,;:"'\\\s])`)
	test := ":es1\n    es2\n    aaes1aa\n    :es3,\n    es1:\n    ees1,\n    ees1 \n    \n    {\n    \"es1 :"
	fmt.Println(match.ReplaceAllString(test, "${1}es4$2"))
}

See the Go demo and the regex demo. Note that the spaces and | chars inside square brackets are meaningful and match these chars literally, thus, you need to remove them all from your pattern.

The regex matches:

• ([,;:"'\\\s])es1\b - Group 1: a comma, or a semi-colon, colon, double or single quotation mark, backslash or whitespace; then es1 as a whole word (\b is a word boundary)
• | - or
• \bes1 - a whole word es1
• ([,;:"'\\\s]) - Group 2: a comma, or a semi-colon, colon, double or single quotation mark, backslash or whitespace