Is this because the go compiler optimized the code?

The issue is not an optimization by the Go compiler, but rather a lack of synchronization. The assignment to i is not followed by any synchronization event, so it is not guaranteed to be observed by any other goroutine. In fact, an aggressive compiler might delete the entire i statement.

The Go Memory Model

The Go memory model specifies the conditions under which reads of a variable in one goroutine can be guaranteed to observe values produced by writes to the same variable in a different goroutine.

To serialize access, protect the data with channel operations or other synchronization primitives such as those in the sync and sync/atomic packages.

If you must read the rest of this document to understand the behavior of your program, you are being too clever. Don't be clever.

Synchronization

In the following example, the assignment to a is not followed by any synchronization event, so it is not guaranteed to be observed by any other goroutine. In fact, an aggressive compiler might delete the entire go statement.

var a string

func hello() {
  go func() { a = "hello" }()
  print(a)
}

The following example shows how to serialize access to i using a sync.Mutex.

package main

import (
    "sync"
    "time"
)

func main() {
    mx := new(sync.Mutex)
    i := 1
    go func() {
        for {
            mx.Lock()
            i  
            mx.Unlock()
        }
    }()